Class Journal - Math 1306 (10162)
What I hear, I forget; what I
see, I remember; what I do, I understand.
- Kung Fu Tzu (Confucius)
One learns the thing by doing the
thing; for though you think you know it,
you
have no certainty until you try. - Sophocles
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Session |
Date |
Read
& Study Section |
Discussion
Topics |
Practice
HW
Problems |
Other
Info |
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29 |
12 - 9 |
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The
Final Exam will be on Wednesday,
December 9, from 10 a.m. - 12:30 p.m., and it will be comprehensive. Bring
a pencil and a scientific calculator to the exam (no cell phone calculators
allowed; no sharing of calculators allowed). MyMathLab homework must
be completed by Tuesday, December 8. You
should review the three semester tests to help prepare for the exam, as well
as your notes and examples from the book. You
may also get additional help for the course in the Math Lab (N925) and from
me (S707). |
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28 |
12 – 2 |
13.1 13.5 |
Section 13.1/13.5 More examples of definite integrals were studied. It is important to remember that a definite integral gives the net area between the curve and the x-axis. In application problems, the units of
measure on a definite integral To help make
practical sense of the integral, intuitively, we can think of the meaning
when f(x) is multiplied by a small amount of x. That is, if f(x)
is the rate of change of electricity usage in million kw/hr
and x is in hours, then f(x)dx
is the amount of electricity used in a small interval of time, and then
summing those up, so that the definite integral calculates the amount of
electricity used over the interval of time from x=a to x=b. |
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27 |
11 - 30 |
13.1 13.5 |
Section 13.1/13.5 A function F(x) is called an antiderivative of a function f(x) if the derivative of F(x) is f(x), that is, F'(x) = f(x). This is the inverse operation of differentiation. The formula for an antiderivative: (1) of a constant c is cx; (2) of xn is xn+1 / (n+1) provided the exponent n is not equal to -1; (3) the constant multiple rule applies to antiderivatives; (4) the sum rule applies to antiderivatives; (5) of ex is ex. Note that antiderivatives are not unique since F(x)=x3 and F(x)= x3 +5 are both antiderivatives of f(x)=3x2. The Fundamental Theorem of Calculus states that |
Section 13.5 # 5 – 13 odd, 19, 21, 23, 25 |
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26 |
11 - 23 |
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Test 3 (12.1 First derivative and graphs; 1.2 Second derivative and graphs; 11.4 General power rule; 11.2/11.4 Derivative of the exponential function and General exponential function derivative rule; Density functions – in packet; 13.4 Definite integral) Bring a pencil and a calculator; no cell phone calculators allowed; no sharing of calculators allowed. All MyMathLab homework for Test
3 must be completed by Sunday, November 22. |
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25 |
11 - 18 |
13.1 |
Section 13.1 In order to calculate the value of a definite integral, we need to learn about antiderivatives. A function F(x) is called an antiderivative of a function f(x) if the derivative of F(x) is f(x), that is, F'(x) = f(x). This is the inverse operation of differentiation. The formula for an antiderivative of a constant c is cx; and the formula for an antiderivative of xn
is xn+1 / (n+1) "add one to the
exponent and divide by that exponent" , provided the exponent n
is not equal to -1. Hence for f(x)=x2
, an antiderivative is F (x) = x3/3.
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24 |
11 - 16 |
13.4 |
Density Function. The properties of a density function are given below. We will estimate the area under a density function curve by counting the number of blocks (whole and fractional) in the area and then multiplying this by the area of one block. Note that the median is a value so that 50% of the population is below this value and 50% of the population is above this value. Practice problems are given in the problem packet “Exponential Functions and Definite Integrals.” Sections 13.4 The definite integral of f(x)
from x=a to x=b equals the net area between the curve y=f(x), the x-axis,
x=a and x=b; it is denoted by Since a definite integral calculates net area, this means
areas above the x-axis are positive and areas below the x-axis are negative.
Study example 3 on page 770. Practice problems are given in the problem
packet “Exponential Functions and Definite Integrals.” |
See below and Section 13.4 # 17 – 27 odd |
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23 |
11 - 11 |
11.2 11.4 |
Sections 11.2/11.4 The exponential function f(x)=ex has a unique property: it is its own derivative, that is, f '(x)=ex. Study example 1 on page 595. The general exponential derivative rule says that a function f(x)=eu(x) has rate of change function f '(x)=eu(x) ∙ u '(x). Study example 6a and 6c on page 617. Density Function. An application of areas under curves involves
density functions. A density function: (1) is never
negative and so its graph is never below the x-axis; (2) the
total area between the graph of the function and the x-axis is 1; (3)
the area between the function curve, the x-axis, x=a and
x=b gives the percent or proportion of the population that is
between x=a and x=b. If the density function is graphed on a grid, we can approximate the
area under the curve for a particular region by counting the number of blocks
(whole and fractional) in the area and then multiplying this by the area of one
block. Class activity 10 was done in class today. |
See below and Problem Packet: Exponential Functions and Definite Integrals # 1 – 9 all [Answers to odd-numbered problems in packet.] |
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22 |
11 - 9 |
11.2 |
Topics/rules related to class activity 9 were reviewed. The general power rule from section 11.4 was reviewed. Section 11.2 A function of the form f(x)=bx is called an exponential function where the base b is a positive number different from 1. Every exponential function of this type has: (1) domain equal to all real numbers; (2) range equal to all positive real numbers; (3) y-intercept equal to 1; (4) the x-axis is a horizontal asymptote; (5) the function is always increasing if b>1, but the function is always decreasing if 0<b<1. The most important exponential function is f(x)=ex where the base e is approximately equal to 2.718, and it is called the exponential function. The exponential function has a unique property: the derivative of the exponential function is itself, that is, if f(x)=ex then f ′(x)=ex. |
Section 11.2 # 1, 5, 17, 21 Section 11.4 # 13, 14, 25, 27, 41, 68, 100
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21 |
11 - 4 |
12.2 11.4 |
Section 12.2 The following are equivalent statements: (1) the graph of a function f(x) is concave upward on an interval; (2) f '(x) is increasing on the interval; (3) f "(x) is positive on the interval. The following are equivalent statements: (1) the graph of a function f(x) is concave downward on an interval; (2) f '(x) is decreasing on the interval; (3) f "(x) is negative on the interval. From these properties, it follows that at an inflection point, the function f(x) is increasing at the fastest rate (or decreasing at the slowest rate) compared to other nearby points – an inflection point is a point of greatest or least slope compared to other nearby points. This is important in application problems. For the function that relates the CPI to time in years: (1) at the inflection point, inflation may be increasing at the fastest rate; (2) if f “(a)>0, inflation is getting worse; (3) if f “(a)<0, inflation is moderating and improving. See problem 5 in the problem packet (Second Derivative and Inflection Points). Section 11.4 The General Power Rule of derivatives is used to find the derivative of a function raised to a power: the derivative of [f(x)]n equals n[f(x)]n-1 ∙f ′(x). Study example 3 on page 613. Class activity 9 was done in class today. |
See below and Section 11.4 # 9, 11, 17 – 23 odd, 29, 31, 37, 29, 43, 45 |
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20 |
11 - 2 |
12.1 12.2 |
Section 12.1 Graph of f(x) versus Graph of f ‘(x). Study exercise 69: use the given graph of y=f ‘(x) to find the intervals on which f(x) is increasing, on which f(x) is decreasing, and the x-coordinates of any local extrema of f(x). Study exercise 75: use the given graph of y=f(x) to find the intervals on which f ‘(x)>0, on which f ‘(x)<0, and the values of x for which f ‘(x)=0. Section 12.2 An inflection point is a point on the graph of a function f(x) where the concavity changes. It is shown that at an inflection f”(x) equals 0 or it does not exist. See page 667. Therefore, to find the inflection points of a function f(x): 1st, we calculate f “(x) and then solve the equation f “(x)=0 to find the possible inflection points; 2nd, we check if the concavity changes at these points. In application problems, an inflection point is a point
where the function is increasing or decreasing at the fastest rate, compared
to other nearby points. An inflection point may be a point of diminishing returns. See page 674. |
Problem Packet: Second Derivative and Inflection Points # 1 – 5 [Answers to odd-numbered problems.] |
All MyMathLab homework assignments for the semester are now open. |
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19 |
10 - 28 |
12.1 12.2 |
Section 12.1 What does the first derivative tell you about the shape of a function’s graph? (a) If f '(x)>0 on an interval, then the function f(x) is increasing on the interval. (b) If f '(x)<0 on an interval, then the function f(x) is decreasing on the interval. (c) If f '(x) changes from negative to positive at a point, then the point is a local minimum (valley point). (d) If f '(x) changes from positive to negative at a point, then the point is a local maximum (peak point). Section 12.2 The second derivative of a function f'(x) is the derivative of the first derivative and it is denoted by f "(x). For example, if f'(x)=x3, then f '(x)= 3x2, and f "(x)=6x. The graph of a function f(x) is concave upward on an interval if f '(x) is increasing on the interval, and the graph is concave downward on an interval if f '(x) is decreasing on the interval. This can be restated by using the second derivative: the graph of a function f(x) is concave upward on an interval if f "(x) is positive on the interval, and the graph is concave downward on an interval if f " (x) is negative on the interval. Class activity 8 was done in class today. |
See below and Text, Page 677 # 1 – 11 odd |
Click here for an
applet that graphs a function f(x) and the line tangent to the graph of f(x)
for a given x-value. Have fun! |
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18 |
10 - 26 |
12.1 |
Problem: Use the rate of change function f '(x): (a) find
the coordinates of any possible turning points; (b) determine if any are local maxima (peak points) or local
minima (valley points) on
the graph. Solution: 1st, solve the equation f '(x) = 0 to find the x-coordinates
of possible turning points, then substitute the x-value into the
original function f(x) to find the matching y-coordinate.
These points are also called critical points
of the function. 2nd, for each possible turning point: calculate the rate of
change f '(x) at a nearby test point on the left side and
at a test point on the right side. 3rd, If the rate of change f '(x) changes from negative to positive, then the function changes from decreasing to increasing and you can conclude that the point is actually a turning point and that it is a local minimum (valley point); If f '(x) changes from positive to negative, then the function changes from increasing to decreasing and you can conclude that the point is actually a turning point and that it is a local maximum (peak point); If f '(x) does not change sign, then the function does not change direction and you can conclude that the point is not a turning point. Recall that the y-intercept of a function f(x) can be found by substituting 0 for x in the function formula f(x); and that the x-intercepts of the function f(x) can be found by substituting 0 for y or 0 for f(x) and then solving the resulting equation. In the graphing window [a,b] x [c,d]:
(1) the x-axis extends from xMin=a
to xMax=b; and (2) the y-axis extends from yMin=c
to yMax=d. Different persons may
choose different graphing windows for the same function. What is important is
whether or not the graphing window that you choose shows the important
properties of the function like the intercepts, turning points, inflection
points, etc. |
Problem Packet: Derivatives and the Shape of the Graph (a) find the coordinates of all possible turning points; (b) determine which are actually turning points; (c) determine which are local maximum (peak point) or local minimum (valley point) ; (d) find the coordinates of all possible inflection points; (e) determine which are actually inflection points; determine a graphing window # 1 – 8 all [Answers to odd-numbered problems in packet.] Text, Page 659 New Directions (a) Find f '(x); (b) Find the (x,y) coordinates of possible turning points; (c) Which of the points in part b are actually turning points? If the point is a turning point, is it a local maximum (peak point) or is it a local minimum (valley point)? Justify your answers. # 19 - 22 all |
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17 |
10 – 21 |
Test 2 (10.1 Introduction to limits; 10.4 The limit definition of the derivative; 10.5 Basic differentiation properties – rules; 10.7 Marginal analysis in business and economics) Bring a pencil and a calculator; no cell phone calculators allowed; no sharing of calculators allowed. All MyMathLab homework for Test
2 must be completed by Tuesday, October 20. |
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16 |
10 – 19 |
10.7 |
Section 10.7 It is helpful to study example 2 on page 571: it involves the price-demand equation, the cost function, the marginal cost function, the marginal revenue function, the break-even points, the marginal profit function, and interpretations. |
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15 |
10 - 14 |
10.7 |
Section 10.7 In economics, the word marginal indicates a rate of change, that is, a derivative. If C(x) is the total cost of producing x items, then C '(x) is called the marginal cost. Similarly, the marginal revenue is the derivative of the revenue function, etc. There is a second interpretation of marginal cost C '(x): it gives the approximate cost of producing the next item, that is, the (x+1)st item. Study example 1 on page 569. Similarly, R '(x) approximates the revenue from the sale of the (x+1)st item, etc. Note in example 2, page 571: we say the equation x=10000-1000p expresses the demand x as a function of the price p; but if the equation is rewritten as p=10-0.001x, we say this equation expresses the price p as a function of the demand x. There is a summary of average cost, average revenue, and average profit on page 574. Study the examples in section 10.7. Class activity 7 was done in class today. |
See below. |
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14 |
10 - 12 |
10.5 10.7 |
Section 10.5 Before using a short-cut differentiation rule, you must check that the problem has the same form as the rule. For example, suppose you want to use the power rule on f(x)=1/x3. First, you need to rewrite the problem to fit the rule: f(x)=1/x3=x-3. Then f '(x)= (x-3)'=-3x-4=-3/x4. Similarly, if f(x)=√x, then f '(x)=(√x)'=(x1/2)'=(1/2)x-1/2=1/(2∙√x). Interpret slope in application problems In application problems, the rate of change of f(x) equals the slope of the tangent line. Also, slope tells you the change in y when x increases by 1 unit; and the units of measure for slope are y-units/x-units. This means if y is in feet and x is in seconds, then slope is in feet per second, which is velocity. Study example 6 on page 558. If y is number of boats sold and x is thousands of advertising dollars, then slope is in boats sold per thousand advertising dollars. So the derivative tells you the increase or decrease in boat sales when advertising spending increases by one thousand dollars. Study problem 83 on page 560. Read ahead in section 10.7. |
Section 10.7 Find and interpret marginal cost or marginal profit or marginal revenue or marginal average cost # 1 – 15 odd |
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13 |
10 - 7 |
10.4 10.5 |
Section 10.4 The derivative f '(a) does not exist when: (1) the graph has a sharp turning point at x=a; (2) the graph has a break at x=a; (3) the graph has a vertical tangent line at x=a. Study Figure 8 on page 549. Section 10.5 There are short-cut differentiation rules that enable us to find the derivative of many functions without having to use to the limit definition. The rules in this section are: the constant function rule on page 553; the power rule on page 554; the constant multiple rule on page 556 and the sum/difference rule on page 557. Carefully study the examples in section 10.5. Class activity 6 was done in class today. |
See below. |
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12 |
10 - 5 |
10.4 |
Section 10.4 The limit definition (4-step process) to find f ‘(a), the derivative of f at a, can also be broken down into these two steps: (1) find the slope of the secant line through the points (x,y)=(a, f(a)) and (x,y)=(a+h, f(a+h)); (2) then find the limit as h approaches 0 of the slope of the secant line (answer from part 1) to get the slope of the tangent line f ‘(a). An example using these two steps was done in class. After you have a formula for f ‘(a), you can substitute for a in this formula to find the slope of the tangent line at x=a. Since the tangent line and the curve appear to coincide at the tangent point, we also say the slope of the tangent line equals the slope of the graph at x=a. Then the geometric information that we learn from the slope of the tangent line can be used to describe the curve at the tangent point. Also, the equation of the tangent line that has slope f ‘(a) and that goes through the point (a, f(a)) can be found by substituting into the slope formula, or into the slope-intercept formula or into the point-slope formula. Sales Application – study example 7 on page 547. Here, S(25) gives total sales after 25 months, but S’(25) gives the rate at which sales are changing after 25 months. |
See below and Section 10.5 Find the indicated derivative # 1 – 51 every other odd Applications # 81, 83, 89 |
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11 |
9 - 30 |
10.4 |
Section 10.4 The following are defined to be equivalent quantities: (1) the instantaneous rate of change of f(x) at x=a; (2) the derivative of f(x) at x=a; (3) the slope of the tangent line to f(x) at x=a; (4) (5) f '(a). Note that the expression in (4) can be described in words as the slope of the secant line through x=a and x=a+h, or it can be described as the average rate of change of f(x) between x=a and x=a+h. Also, when the directions state to use the “4-step method” to find f ‘(a), this means to calculate f ‘(a) by using the limit definition. Carefully study examples 3 and 4 this section. Class activity 5 was done in class today. |
Section 10.4 Find slope of secant
line/tangent line # 1, 27 Find f’(x) by using the
limit definition # 5 – 13 odd, 19, 25, 39 Determine where f ‘(x)
exists # 31 – 38 Applications # 59, 60, 61 |
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10 |
9 - 28 |
10.1 |
Section 10.1 Let’s review the solution method to estimate f ‘(a) from a table. We choose several values of h; each h gets closer and closer to 0. Then for each h, we compute the average rate of change f[a,a+h]. Finally, we study the table: as h approaches 0, do the values of the average rate of change approach a definite number? If yes, then this definite number is an approximation of f '(a). This process can be described more formally by using the concept of a limit. Function Limit:
If the value of f(x) can be made arbitrarily close to L by taking x
sufficiently close to a, but not equal to a, then we write Direct Substitution
Property Let x=a be in the domain of the function: (1) if f(x) is a polynomial function or a rational function then When the direct substitution property can be used, we say the function is continuous at x=a. Study example 6 on page 506. Direct Substitution Property Cannot Be Used: If f(a) is not defined so that x=a is not in the domain of f(x) then the direct substitution property cannot be used. However, it is sometimes possible to use algebra to show that the formula for f(x) equals the formula for a different function g(x) for all x except x=a. In this case, if the limit for g(x) as x approaches a exists, then the limit for f(x) will also exist and have the same value. This is because the limit definition uses x values close to a but not equal to a. Study example 8a on page 507 and example 9 on page 509. Tangent Line The
definition of the tangent line to the curve y=f(x) at the point (x,y)=(a, f(a)) is the line
through this point with slope equal to
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Section 10.1 Find the limit # 49, 50,
51, 52, 53, 54, 55, 57 |
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9 |
9 - 23 |
2.1 2.3 2.4 10.4 Class notes |
Test 1 (Function properties, including function notation, domain and range, intercepts, turning points, increasing on an interval, decreasing on an interval; profit-loss analysis including profit-demand function, cost function, revenue function, profit function and average cost function, break-even points; average rate of change of a function from a graph and from a formula; instantaneous rate of change of a function from a graph and from a table; interpret values in the context of the problem.) Bring a pencil and a calculator; no cell phone calculators allowed; no sharing of calculators allowed. All MyMathLab homework for Test 1 must be completed by Tuesday, September 22. |
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8 |
9 – 21 |
Problems related to test 1 topics were discussed. |
See below. |
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7 |
9 - 16 |
10.4 Class notes |
Tangent Line
Approximation to f '(a) The instantaneous rate of change of f(x) at x=a
is written as Numerical (Use a Table) Approximation of f '(a) The method of zooming in on a point on a smooth unbroken graph until the graph appears to be a straight line, and then choosing 2 points on this line to approximate f '(a) can be carried out with a function formula. The two points are (x,y)=(a, f(a)) and (x,y)=(a+h, f(a+h)) where the increment h is chosen to be a small number, close to 0. Then the average rate of change of f(x) between these two points, which equals the slope, is calculated. The process of zooming in is replaced by calculating the average rate of change for several values of h that approach 0, for example, we can use h=0.5, h=.25, h=0.1, h=0.01, and h=0.001. These calculations are set up in a table as shown in the packet of problems “Numerical Approximation of f '(a).” We then study the completed table: as h approaches 0, do values of the average rate of change approach a definite number? If yes, then this definite number is an approximation of f '(a). Problem 1 in the packet was discussed in class. Class activity 4 was done in class today. |
See below and Numerical Approximation of f '(a) # 1 - 5 |
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6 |
9 - 14 |
10.4 Class notes |
Average Rate of Change of a Linear Function Motivating example: Think of Michelle with her cruise control set at 60 mph: (1) the graph relating distance traveled to elapsed time is a straight line with equation f(x)=60x, where 60 equals the slope; (2) her average velocity is 60 mph between any two times; and (3) her velocity at any instant is 60 mph. In summary, if the function f is a linear function (graph is a straight line) with formula f(x)=mx+b, then: (1) the average rate of change of f between x=a and x=b always equals the slope m, that is f[a,b] =m for any two points x=a and x=b; (2) the instantaneous rate of change of f at x=a, which is denoted f '(a), always equals the slope m. Graph Approximation of f '(a) For a smooth unbroken curve, when you zoom-in at the point where x = a, it will appear to be a straight line and we will assume the curve has the same properties as the line. Therefore, the instantaneous rate of change of the function at x = a is the slope of this line, approximately. So we can choose two points on this “line” and calculate the slope to get an approximation of f '(a). As it turns out, if you could repeatedly zoom-in
indefinitely, the graph will look more and more like a particular line called
the tangent line at the point x = a. In fact, |
Section 10.4 Use a graph to find the slope of tangent line to find f '(1), f '(2) and f '(3) # 9, 11, 13, 15, 17, 19, 21 Use the graph to find the slope of the tangent line # 27d Sketch a tangent line, find the slope of tangent line, interpret the slope of tangent line # 1 - 11 |
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5 |
9 - 9 |
10.4 Class notes |
Average Rate of Change If we take a car trip and compute the average velocity to be 60 mph, we are also computing the average rate of change in distance traveled with respect to elapsed time. The average rate of change of the function f(x) between x=a and x=b is denoted by f[a,b] and the formula is f[a,b] = [ f(b) - f(a) ] / [b - a ]. The units of measure for the average rate of change are "output units per input unit". The interpretation of a positive average rate of change is: "On average, from x=a to x=b, the function f(x) increases by f[a,b] output units per unit increase in the input." If the average rate of change is negative, then we replace "increases" by "decreases." Note that the average rate of change of the function f(x) between x=a and x=b is the same as the slope of the line that goes through the graph of f(x) at the points (x1,y1)=(a, f(a)) and (x2,y2)=(b, f(b)); this line is called a secant line. Hence many properties of average rate of change can be determined by visual inspection of the graph of f(x): (1) a line that rises from left to right has positive slope; (2) a line that falls from left to right has negative slope; (3) a horizontal line has zero slope; (4) a vertical line has no slope. The average rate of change can be computed from a graph, a formula or a table. Class activity 3 was done in class today. |
Section 10.4 Find the average rate of change (slope of secant line) # 3ab, 4ab, 29a, 30a, 59a, 60a Find the average rate of change and interpret # 1 – 4 |
There is an Average Rate of Change homework posted in MyMathLab. |
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4 |
9 - 2 |
2.3 Class notes |
Section 2.3 To find the break-even points, you will solve an equation: solve “profit = 0.” The break-even points in problem 62, page 93, were found in class by using the quadratic formula. Another function used in profit-loss analysis is the average cost function For example, if the cost to produce x items is C(x)=5000+0.5x2, then the average cost per item is
More Function Properties A turning point on the graph of a function is a point on the graph where the graph changes direction, from increasing to decreasing or from decreasing to increasing. For example, see the graph in example 1 on page 47 in section 2.1. There is a turning point at the point with x=0; this point is also the vertex of this parabola. When a graph rises from left to right, we say the function is increasing on the matching interval of x-values. When a graph falls from left to right, we say the function is decreasing on the matching interval of x-values. Some texts include the endpoints of the interval, and other texts do not include the endpoints. For the graph in example 1 on page 47: (1) the function is increasing on the interval where x<0 or in interval notation (-infinity,0), and (2) the function is decreasing on the interval where x>0 or in interval notation (0,infinity). |
See below and Properties of Functions: Find the turning points; intervals where the function is increasing; intervals where the function is decreasing # 1 – 5 |
There is an Properties of Functions homework posted in MyMathLab. |
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3 |
8 - 31 |
2.1 2.3 |
Section 2.1 Profit-Loss Analysis The cost function C(x) gives the cost in dollars to produce x items. The cost function includes fixed costs and variable costs. The revenue function R(x) gives the revenue in dollars from selling x items. The profit function P(x) gives the profit from selling x items. Also, profit = (revenue – cost): (1) if profit is negative, then revenue is less than cost and the company has a loss; (2) if profit is positive, then revenue is greater than cost and the company makes a profit; (3) if profit equals zero, the company will break-even. In addition, in price-demand functions, x is the number of items that can be sold at p dollars per item. It is important to pay attention to the units of measure in application problems. Study example 7, page 56, and exercises 117, 119, and 121 on pages 61-62 in the text. Section 2.3 To find the break-even points, you will solve an equation: solve “profit = 0.” See example 4, page 84; the quadratic formula is used in this example. Class activity 2 was done in class today. |
See below and Section 2.1 Price-demand # 117, 119, 121 Section 2.3 Break-even points # 61c, 62c |
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2 |
8 - 26 |
2.1 |
Function values from a graph. For the function f(x), if a point (x,y) is on the graph, then x=input and y=output and we write y=f(x). Given the graph of a function f(x), two typical questions are: (1) find the value of y if y=f(-5); (2) find the value of x if 0=f(x). See problems 45 and 49 on page 60. Finding intercepts for a function. For the function y=f(x), we substitute 0 for y to find the x-intercepts and we substitute 0 for x to find the y-intercepts. To find the x-intercepts of f(x)=2x-1, you must solve the linear equation 0=2x-1. To find the x-intercepts of f(x)=x2+5x+4, you must solve the quadratic equation 0= x2+5x+4; this can be done by factoring or using the quadratic formula. Finding the domain of a function formula. If the domain is not given with a function formula, then the domain is assumed to be the largest possible set of inputs. However, each number in the domain must have a matching output that is a real number. (1) If the function formula has a variable in a denominator, then any number that makes the denominator equal zero must be omitted. For example, f(x)=12/(x-5) has domain all real numbers except 5. The rule here is to solve the “denominator = 0.” (2) If the function formula has a variable under a square root, then only numbers that make the radicand equal zero or greater are included in the domain. For example f(x)=√(x-4) has domain x>4. The rule here is to solve the “radicand > 0.” Finding the domain
of a function graph. To find the domain of a function graph, find
vertical lines that bound the graph. Then use the x-intercepts of these
vertical lines to set up the domain. To find the range of a function graph,
find horizontal lines that bound the graph. Then use the y-intercepts of
these horizontal lines to set up the range. See the graphs on page 65. |
See below |
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1 |
8 - 24 |
2.1 |
A function matches each possible input with exactly one output; but repeated outputs are OK. There are different forms for a function, including, a table, a graph and a formula. A formula is often expressed by using function notation. For example, the function formula f(x)=2x-1 has input variable x. To find the matching output for the input x=0, we substitute 0 for x all the way across in the formula and get f(0)=2(0)-1=-1. To graph the function given by a formula, we let y be the output, and so we write, y=f(x). On a function graph, it is customary for the x-axis to be the input axis and for the y-axis to be the output axis. If a point (x,y) is on the graph, then x=input and y=output and so we write (x,y)=(input,output). A point (x,y) is on the x-axis if y=0; and so the rule to find the x-intercepts is to substitute 0 for y in the function formula and then solve for x. A point (x,y) is on the y-axis if x=0; and so the rule to find the y-intercept is to substitute 0 for x in the function formula, and then simplify. |
Section 2.1 Function notation # 45, 47, 49,51, 53-58 all Find the domain # 65-69 all Price-demand # 117, 119, 121 Review of Functions # 1 - 8 |
You must register in MyMathLab during the first two weeks of the semester (by September 6). If you attend one of the three Algebra Review Sessions to be held in the Math Lab (N925) on Wednesday (8-26) from 1-2:15 pm, Thursday (8-27) from 3:30-4:45 pm or Saturday (8-29) from 12-1:15 pm, then you will have 10 points added to your score on the Algebra Review assignment in MML. Class Activity # 1: Register in MyMathLab and in the UHD Math Portal during the first week of classes (by Aug. 30). MML Algebra Review: If you score 80% or higher on the Algebra
Review that is posted in the homework assignments of MML, you will earn 5
Bonus Points on Test 1 (3 attempts per exercise; due date is Tue. 9/21). |
