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Math 170 Calculus I
Exam I Review Solutions
1. Find the domain of each of the following functions:
2x + 1
x2 − 7x + 12
The function is undefined where the denominator is equal to 0, so find those values:
(a) f (x) =
0 = x2 − 7x + 12 = (x − 3)(x − 4)
when x = 3 and x = 4. The domain includes all real numbers except these values:
[
[
(−∞, 3) (3, 4) (4, ∞)
√
(b) (b) f (x) = x2 − 4
This function is only defined where x2 − 4 ≥ 0, so when x2 ≥ 4. Then x could be any value less than or
equal to −2, or any value greater than or equal to 2:
[
(−∞, −2) (2, ∞)
(c) (c) f (x) = 8x3 + 3x − 1
This is a polynomial, so it is defined everywhere. The domain is all real numbers.
(d) (d) f (x) = cos(x + 1)
Since cos(x) is defined for all values of x, simply adding 1 to x first will not cause the function to be
undefined anywhere. The domain is again all real numbers. Another way to see this is to view f (x) as
the composition of two functions (cos(x) and x + 1) that are both defined everywhere, so its domain is
their intersection of their domains, which is everything.
2. If y = x2 + 7x − 2, find any values of x where y = −14. (Hint: there are two real solutions.)
−14 = x2 + 7x − 2
0 = x2 + 7x + 12 = (x + 3)(x + 4)
x = −3, −4
3. Find (f ◦ g)(x) and (g ◦ f )(x) if f (x) = 3x and g(x) = 2x + 5
(f ◦ g)(x) = f (g(x)) = f (2x + 5) = 3(2x + 5) = 6x + 15
(g ◦ f )(x) = g(f (x)) = g(3x) = 2(3x) + 5 = 6x + 5
4. If f (x) = 2(x − 1)3 + 4, find functions g and h so that f = g ◦ h.
There are several options here; one possible choice is g(x) = 2x3 + 4 and h(x) = x − 1
5. Decide if the function f (x) = 6x − 8 has an inverse, and if so, find it. Verify that you have found the inverse
of f (x). This function does have an inverse.
y = 6x − 8
x = 6y − 8
6y = x + 8
y = f −1 (x) =
x+8
6
To check, show that BOTH compositions produce x:
(f ◦ f −1 )(x) = 6
x+8
6
(f −1 ◦ f )(x) =
!
− 8 = (x + 8) − 8 = x
(6x − 8) + 8 6x
=
=x
6
6
x2 − 3x
2x − 2
Find any values where the denominator is 0. If this value also makes the numerator 0, then you do not have
an asymptote. Otherwise, you do.
2x − 2 = 0
6. Find the vertical asymptote(s), if any, of the function f (x) =
when x = 1. Since (1)2 − 3(1) = −2, then we do have a vertical asymptote at x = 1.
7. Refer to the graph of the function f (x) below:
(a) Find f (−2) = -1, f (0) = 0, f (1) = 1, and f (4) = 3
(b) Find lim f (x) DNE,
x→−2
lim f (x) = -4, lim f (x) = 2, lim f (x) = 0, and lim f (x) = 4
x→−2+
x→1
x→0
x→2−
(c) Find any points x where is f (x) discontinuous, and classify the type of discontinuity x = −2, 1, 2 (don’t
worry about classifying them yet; sorry!)
(d) Find the roots of f (x) x = −3, 0, 3
8. Find the limits:
(a) lim 2x2 − 3x + 5 = 2(−1)2 − 3(−1) + 5 = 10
x→−1
(b) lim
1
= +∞
|x − 4|
(c) lim
4x − 12
4(x − 3)
4
4
4
= lim
= lim
=
=
− x − 6 x→3 (x − 3)(x + 2) x→3 x + 2 3 + 2 5
x→4−
x→3 x2
(x2 − 1)(x2 + 1)
(x − 1)(x + 1)(x2 + 1)
x4 − 1
= lim
= lim
= lim (x + 1)(x2 + 1)
x−1
x−1
x→1+
x→1+
x→1+
x→1+ x − 1
= (1 + 1)(12 + 1) = 4
(d) lim


x − 3,
9. Let f (x) = x2 ,


3x + 6,
x < −2
−2 ≤ x ≤ 5
x>5
(a) Find f (0) = 02 = 0, f (−6) = −6 − 3 = −9, and f (2) = 22 = 4
(b) Find
lim f (x) = lim x − 3 = −2 − 3 = −5
x→−2−
x→−2−
(c) Find lim f (x) DNE, since
x→−2
lim f (x) = 4 6= lim f (x)
x→−2+
x→−2−
(d) Find lim f (x) DNE, since the limit from the left is 25 and the limit from the right is 21
x→5