Download SOLUTIONS TO Practice Exam #1 1. Graph the function h(x) = √ x +

SOLUTIONS TO
Practice Exam #1
1. Graph the function h(x) =
√
x + 3 using the techniques of shifting, compressing, stretching
and/or reflecting. Start with the graph of the basic function and show all stages.
In Example 10 on page 249 you’ll find the graph of y =
√
x. The graph of h is obtained from
that by shifting over 3 units to the left (so the graph starts at the point (−3, 0)).
2. A semicircle of radius r is inscribed in a rectangle so that the diameter of the semicircle is
the length of the rectangle. Express the perimeter p of the rectangle as a function of r.
You hopefully were able to draw a picture yourself. If not, you can see one in excercise 23 on
page 271. Once you have the picture, you can see that the length of the rectangle is 2r and the
height is r. The perimeter p is then 2r + r + 2r + r, i.e., p(r) = 6r.
3. For the function f (x) = x2 − 9, x ≥ 0, find f −1 . State the range of f . Also, graph f and
f −1 together on the same coordinate axes.
Start with the equation y = x2 − 9 and solve for x. You get
y + 9 = x2
√
p
y + 9 = x2
p
y + 9 = |x|.
√
Here is where you use x ≥ 0 to get y + 9 = x. Replacing y with x gives you the formula for the
√
inverse function: f −1 (x) = x + 9. The range of f is the domain of f −1 . For this, you must not
take the square root of a negative number, so x + 9 ≥ 0, x ≥ −9. The range of f is all numbers
greater than or equal to −9.
Sorry, putting graphs in these pdf files is a real pain so you’re on your own there. Remember
that the graph of f −1 is the graph of f reflected about the line y = x. Also, be sure to label axes
and scales on your graphs!
4. Start with the graph of y = ex and use transformations to graph the function f (x) = −e1−x .
Determine the domain, range and horizontal asymptote of f .
1
Start with the graph of y = ex and reflect it about the y-axis to get the graph of y = e−x .
Shift this graph one unit to the right to get the graph of e1−x . Finally, reflect this about the x-axis
to get the graph of f (x).
You can exponentiate any number, so the domain of f is all numbers (which you can also see
from the graph). From the graph, you should be able to see that the range of f is all negative
numbers and the x-axis is a horizontal asymptote to the graph.
5. Find the domain of the function f (x) = log2 (x2 − 4).
Here your only concern is to make sure what’s inside the logarithm is positive, so you need
x2 − 4 > 0. You can factor to get (x − 2)(x + 2) > 0. This will happen if either both factors are
positive or both factors are negative. Both x − 2 and x + 2 will be positive if x > 2. Both will be
negative if x < −2. Thus, the domain of f is all numbers greater than 2 and all numbers less than
−2; in interval notation (−∞, −2) ∪ (2, ∞). (The ∪ means “union”.)
6. Express y as a function of x if ln y = ln C − 4x. The constant C is a positive number.
Exponentiate both sides (base e) to get y out of the logarithm:
eln y = eln C−4x
y = eln C /e4x
y = C/e4x .
7. The population of a country follows the exponential law. If the population increased from
1.2 million to 2 million from 1950 to 1990, what will the population be in 2010?
Let P (t) be the population as a function of time t in years after 1950. Then P (t) = P0 ekt .
You’re told the population is 1.2 million in 1950, so that’s P0 : P (t) = 1, 200, 000ekt . Next, you
need to find k. To do this you use the population in 1990, 40 years after 1950, is 2 million:
P (40) = 1, 200, 000ek40 = 2, 000, 000.
2
It’s important that you put in 40 for t, not 1990. Solving this equation for k gives you
1, 200, 000e40k = 2, 000, 000
e40k =
2, 000, 000
20
=
= 5/3
1, 200, 000
12
ln(e40k ) = ln(5/3)
40k = ln(5/3)
k=
ln(5/3)
.
40
Once you have k, you can find the population in 2010, which is 60 years after 1950, by plugging in
t = 60:
P (60) = 1, 200, 000e60k = 1, 200, 000e60 ln(5/3)/40 = 1, 200, 000e3 ln(5/3)/2 .
3