Download Math 113 Homework Solutions: Linear Cost Function and Graphing Sinusoidal Functions and more Exercises Calculus in PDF only on Docsity!
Math 113 HW #2 Solutions
- Exercise 1.2.18. The monthly cost of driving a car depneds on the number of miles driven. Lynn found that in may it cost her $380 to drive 480 mi and in June it cost her $460 to drive 800 mi.
(a) Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model. Answer: Since weโre using a linear model, we want, first, to find the slope of the line. Clearly, if the line goes through the points (480, 380) and (800, 460), then the slope of the line is 460 โ 380 800 โ 480
Now we use the point-slope formula with the point (480, 380):
C โ 380 =
(d โ 480) = d 4
Hence, C(d) = d 4
(b) Use part (a) to predict the cost of driving 1500 miles per month. Answer: Plugging in d = 1500 to the expression for C(d) yields
C(1500) =
so we estimate that it will cost $635 to drive 1500 miles per month. (c) Draw the graph of the linear function. What does the slope represent? Answer: The graph is shown in Figure 1. The slope represents the marginal cost of driving an additional mile.
0 100 200 300 400 500 600 700 800 900 1000
100
200
300
400
500
Figure 1: The cost function C(d) = d 4 + 260.
(d) What does the y-intercept represent? Answer: The y-intercept, $260, gives the cost of owning a car which is independent of the number of miles driven (for example, the cost of insurance would be included in this cost). An economist might call this the โfixed cost of drivingโ. (e) Why does a linear function give a suitable model in this situation? Answer: A linear function gives a suitable model because we would expect the cost of driving to be more or less proportional to the number of miles driven.
- Exercise 1.3.14. Graph the function y = 4 sin 3x by hand, not by plotting points, but by starting with the graph of one of the standard functions given in Section 1.2, and then applying the appropriate translations. Answer: This graph will be like the graph of f (x) = sin x, except stretched vertically by a factor of 4 and compressed horizontally by a factor of 3, yielding the graph shown in Figure 2.
-5 -2.5 0 2.5 5
2
4
Figure 2: The graph y = 4 sin 3x
- Exercise 1.3.26. A variable star is one whose brightness alternately increases and decreases. For the most visible variable star, Delta Cephei, the time between periods of maximum bright- ness is 5.4 days, the average brightness (or magnitude) of the star is 4.0, and its brightness varies by ยฑ 0 .35 magnitude. Find a function that models the brightness of Delta Cephei as a function of time. Answer: Since the brightness alternately increases and decreases, we should model it by a periodic function, which suggests either the sine or the cosine. Letโs try using sine. Then we
(c) f โฆ f Answer: By definition,
(f โฆ f )(x) = f (f (x)) = f
x 1 + x
x 1+x 1 + (^) 1+xx
x 1+x 1+x 1+x +^
x 1+x
x 1+x 1+2x 1+x
x 1 + 2x
This function is well-defined except when f is not well-defined (which happens when x = โ1) and when the denominator equals zero, so the domain of f โฆ f is the set {x โ R : x 6 = โ 1 / 2 , x 6 = โ 1 }.
(d) g โฆ g Answer: By definition, (g โฆ g)(x) = g(g(x)) = g (sin 2x) = sin (sin 2x). Since the sine function is defined on all real numbers, the domain of g โฆ g is all of R.
- Exercise 1.3.44. Express the function
G(x) = 3
x 1 + x in the form f โฆ g. Answer: Define the functions f (x) = 3
x g(x) =
x 1 + x
Then f โฆ g(x) = f (g(x)) = f
x 1 + x
x 1 + x
= G(x),
as desired.
- Exercise 1.3.52. Use the given graphs of f and g to estimate the value of f (g(x)) for x = โ 5 , โ 4 , โ 3 ,... , 5. Use these estimates to sketch a rough graph of f โฆ g. Answer: To determine the value of f (g(x)) when x = 0, we estimate from the graph that g(0) โ 2 .8 and f (2.8) โ โ 0 .5. Thus, f (g(0)) โ f (2.8) โ โ 0. 5. Similar calculations yield the following values of f (g(x)), with a possible graph of f (g(x)) shown to the right: x g(x) f (g(x)) โ 5 โ 0. 2 โ 4 โ 4 1. 2 โ 3. 3 โ 3 2. 2 โ 1. 7 โ 2 2. 8 โ 0. 5 โ 1 3 โ 0. 2 0 2. 8 โ 0. 5
x g(x) f (g(x)) 1 2. 2 โ 1. 7 2 1. 2 โ 3. 3 3 โ 0. 2 โ 4 4 โ 1. 9 โ 2. 2 5 โ 4. 1 1. 9
-5 -4 -3 -2 -1 0 1 2 3 4 5
2
4
- Exercise 1.3.56. An airplane is flying at a speed of 350 mi/h at an altitude of one mile and passes directly over a radar station at time t = 0. (a) Express the horizontal distance d (in miles) that the plane has flown as a function of t. Answer: Since the plane is traveling at a constant speed of 350 mi/h and since distance equals rate times time, we have that d(t) = 350t is the horizontal distance function. (b) Express the distance s between the plane and the radar station as a function of d. Answer: Using the Pythagorean Theorem, the distance s from the plane to the radar
d
s
Figure 3: Schematic picture of airplane and radar station
station is given by s^2 = 1 + d^2 , so we have that s(d) =
1 + d^2. (c) Use composition to express s as a function of t. Answer: If we want to express the distance from the radar station as a function of t, we simply take the composition s โฆ d(t) = s(d(t)) = s(350t) =
1 + (350t)^2 =
1 + 122, 500 t^2.
- Exercise 1.3.62. If f (x) = x + 4 and h(x) = 4x โ 1, find a function g such that g โฆ f = h.
Answer: At first glance, it seems like we could just multiply the results of f by 4 to get h. Of course, this doesnโt work, since 4(x + 4) = 4x + 16 6 = 4x โ 1. However, if we subtract 17 after multiplying by 4, we will get the right answer. In other words, if we define g(x) = 4x โ 17 , then we see that (g โฆ f )(x) = g(f (x)) = g(x + 4) = 4(x + 4) โ 17 = 4x โ 1 = h(x), as desired.
