Download Math 170 Calculus I Exam II Review Solutions 1. Find the vertical

Math 170 Calculus I
Exam II Review Solutions
1. Find the vertical and horizontal asymptotes, if any, of the function f (x) =
x2 − 3x
2x − 2
There’s a vertical asymptote at x = 1, because the denominator is 0 there and the numerator is
not.
To find a horizontal asymptote (if there is one), take the limit as x = ∞:
x2 − 3x
=∞
x→∞ 2x − 2
lim
Since this is not a number, there is no horizontal asymptote.
2. Find the limits:
4x − 12
(a) lim 2
x→3 x − x − 6
lim
4x − 12
4(x − 3)
4
4
= lim
= lim
=
− x − 6 x→3 (x − 3)(x + 2) x→3 x + 2 5
x→3 x2
(b) lim
x→0+
1
+3
x2
lim
x→0+
1
+3=∞
x2
7x2 + 2x − 3
(c) lim
x→∞ 4x2 + 5x + 11
7x2 + 2x − 3
7
=
x→∞ 4x2 + 5x + 11
4
lim
(d) lim
x→0
sin(4x)
5x
lim
x→0
sin(4x) 4
=
5x
5
3. Refer to the graph of the function f (x) below to find any points x where is f (x) discontinuous, and
classify the type of discontinuity:
x = −2 and x = 2 are jump discontinuities, x = 1 is a removable, and there do not appear to be
any infinite discontinuities (maybe near x = 4, but it’s hard to see).
4. Where is the function f (x) =
x2 − 9
discontinuous?
x2 + 8x + 15
The function is not continuous anywhere it is undefined, so just find any values of x where
x2 + 8x + 15 = 0. You can factor this one:
x2 + 8x + 15 = (x + 3)(x + 5) = 0
when x = −3 and x = −5.
This was not part of the question, but note that you can simplify the function to one that is
defined at x = −3, since the numerator also has a factor of x + 3. This means that x = −3 is a
removable discontinuity, and x = −5 is an infinite discontinuity (aka vertical asymptote).
5. Show that the function f (x) = 3x2 − 5x − 6 has a root between -1 and 0. What theorem did you
use?
Since
f (−1) = 3(−1)2 − 5(−1) − 6 = 2 > 0
and
f (0) = 3(0)2 − 5(0) − 6 = −6 < 0
and f is continuous, then there must be some point between -1 and 0 where f crosses the x-axos
(ie, has a root). This is using the Intermediate Value Theorem.
6. Which of the following functions are continuous p
everywhere?
3
2
(a) sin(x + 2x)
(b) 3x + tan(x)
(c) (5x + 3)
The first function is the only one. It’s a composition of two functions that are continuous
everywhere (for all real numbers). The second has a discontinuity where tan(x) is undefined.
The third function is not defined for any values of x less than −3/5.
7. Find a point on the graph of y = x2 where the tangent line to y is parallel to the line y = 6x − 4.
Parallel lines have the same slope, so the tangent line to the curve y = x2 at the x we’re looking
for has slope 6. This means the derivative of the curve at this x is 6, and since the derivative is
2x, then x = 3.
8. If f (x) = 3x2 − 5, find the equation of the tangent line to f (x) at x = 2.
f 0 (x) = 6x, so the slope of the tangent line at x = 2 is f 0 (2) = 6(2) = 12. The tangent line
must also shared a point with f (x) when x = 2, which means it passes through the point
(2, f (2)) = (2, 3(2)2 − 5) = (2, 7)
So in point-slope form, the tangent line has equation
y − 7 = 12(x − 2)
9. Find f 0 (x) using the limit definition of the derivative when f (x) = 3x2 − 1. Check your answer using
the power rule.
3(x + h)2 − 1 − (3x2 − 1)
3(x2 + 2xh + h2 ) − 1 − (3x2 − 1)
= lim
h→0
h→0
h
h
2
6xh + 3h
= lim
= lim (6x + 3h) = 6x
h→0
h→0
h
lim
10. Find f 0 (x), using any method you prefer.
√
(a) f (x) = 2x + 1(5x + 2)
Use the product rule, then the chain rule as needed:
√
√
d √
d
1
[ 2x + 1](5x + 2) + ( 2x + 1) [5x + 2] = (2x + 1)−1/2 (2)(5x + 2) + 2x + 1(5)
dx
dx
2
√
= (2x + 1)−1/2 (5x + 2) + 2x + 1(5)
f 0 (x) =
(b) f (x) =
1 + 4x3 − 8x5 + x7
x2
Rewrite as powers of x, then use the power rule:
f (x) =
1 + 4x3 − 8x5 + x7
= x−2 + 4x − 8x3 + x5
x2
f 0 (x) = −2x−3 + 4 − 24x2 + 5x4
1
√
(c) f (x) = x6 − 4 x + 3
4x
Rewrite as powers of x, then use the power rule:
1
f (x) = x6 − 4x1/2 + x−3
4
3
f 0 (x) = 6x5 − 2x−1/2 − x−4
4
Note that if you write the last term as (4x3 )−1 , you could use the chain rule to find its
derivative, but the power rule is easier.
(d) f (x) = π e − eπ + ln(1)
Everything here is constant, so f 0 (x) = 0.
(e) f (x) =
x2 − 2
2x2 + 1
!3
Use the chain rule, and the quotient rule. No need to simplify:
0
f (x) = 3
x2 − 2
2x2 + 1
2
(2x2 + 1)(2x) − (x2 − 2)(4x)
(2x2 + 1)2
(f) f (x) = sin(x3 + 2)
Use the chain rule:
f 0 (x) = cos(x3 + 2)(3x2 + 0) = 3x2 cos(x3 + 2)
(g) f (x) =
sin(x)
cos(x)
!4
We could use the chain rule, along with the quotient rule. But it’s easier if we recognize
that sin(x)/ cos(x) = tan(x), so y = tan4 (x). Now we just need the chain rule, and the
derivative of tan(x):
y 0 = 4 tan3 (x) sec2 (x)
(h) f (x) =
− 2x3 − 5x
8x2 − 10x
Simplify first:
−2x2 − 5
8x − 10
Now use the quotient rule, and don’t worry about simplifying:
y=
y0 =
(8x − 10)(−4x) − (−2x2 − 5)(8)
(8x − 10)2