Math 201
Final Exam Key
Section 3
Spring 2013
1-2. Consider
f (x) :=
x3
− x2
3
on the interval [−3, 3].
(a) Find f 0 (x).
Solution. f 0 (x) = x2 − 2x.
(b) Find f 00 (x).
Solution. f 00 (x) = 2x − 2.
(c) Use (a) and (b) to find all local and global maximum and minimum values of
the function over the interval [-3, 3]. Do not forget to also check the
endpoints.
Solution. Using (a) we set f 0 (x) = x2 − 2x = 0 =⇒ x = 0 and x = 2. Using
(b), we get f 00 (0) = −2 < 0 =⇒ f (0) = 0 is a local maximum;
f 00 (2) = 2 > 0 =⇒ f (2) = − 43 is a local minimum. We now check endpoints:
f (−3) = −18, and f (3) = 0. We thus conclude that the minimum occurs at
f (−3) = −18, and the maximum at f (0) = 0 = f (3).
(d) Use (c) to find the intervals where f is increasing.
Solution. By (c) f increases on [−3, 0), and on (2, 3]
(e) Use (b) to find the points of inflection of f over the interval [-3, 3].
Solution. Using (b) we set f 00 (x) = 2x − 2 = 0 =⇒ x = 1. Since f (1) = − 23 ,
the only point of inflection of f is 1, − 23 .
(f) Use (c) and (e) to find the intervals where f is concave down.
Solution. By (c) we get that f is concave up on (1, 3] since 2, − 34 is a local
minimum. Thus f is concave down on [−3, 1).
3-4. The following is the graph of the derivative f 0 of a function f . Accurately answer
the following questions:
y
5
0
6
-
x
(a) Find the the interval(s) where the function f is increasing.
Solution. f is increasing when f 0 > 0. According to the graph, this happens
approximately between −2.5 and 0, and if x > 2.5.
(b) Find the the interval(s) where the function f is concave down.
Solution. f is concave down when f 0 is decreasing. According to the graph,
this happens approximately if −1.2 < x < 1.2.
(c) Find the x-coordinate(s) of the points where the function f has a local
maximum.
Solution. f has a local maximum if f 0 changes from positive to negative.
According to the graph, this happens at x = 0.
(d) Find the x-coordinate(s) of the points where the function f has a local
minimum.
Solution. f has a local minimum if f 0 changes from negative to positive.
According to the graph, this happens approximately at x = ±2.5.
(e) Find the x-coordinate(s) of the inflection points of the function f .
Solution. These are the points where concavity changes, i.e., where the
derivative changes from increasing to decreasing or vice versa, i.e., at the
points where f 0 has local maxima or minima. According to the graph, this
happens approximately at x = ±1.2.
5. Suppose f 0 (2) = 5, g0(2) = 2, f (2) = −2 and g(2) = 3, f 0 (3) = 6 and g0(−2) = 3.
Find the derivative at x = 2 of each of the following functions:
(a) s(x) = f (x) − g(x).
Solution. s0 (x) = f 0 (x) − g 0 (x), hence
s0 (2) = f 0 (2) − g 0 (2)
= 5−2
= 3.
(b) p(x) = f (x)g(x).
Solution. p0 (x) = f (x)g 0 (x) + f 0 (x)g(x), hence
p0 (2) = f (2)g 0 (2) + f 0 (2)g(2)
= (−2)(2) + 5(3)
= 11.
(c) q1 (x) =
f (x)
.
g(x)
Solution.
q10 (x)
f 0 (x)g(x) − f (x)g 0 (x)
=
, hence
(g(x))2
f 0 (2)g(2) − f (2)g 0 (2)
(g(2))2
5(3) − (−2)(2)
=
(3)2
19
=
.
9
q10 (2) =
(d) h(x) = f (g(x)).
Solution. By the Chain rule we have h0 (x) = (f (g(x)))0 = f 0 (g(x))g 0 (x). Hence
h0 (2) = f 0 (g(2))g 0 (2) = f 0 (3)(2) = (6)(2) = 12.
(e) k(x) = g(f (x)).
Solution. By the Chain rule we have k 0 (x) = (g(f (x)))0 = g 0 (f (x))f 0 (x).
Hence k 0 (2) = g 0 (f (2))f 0 (2) = g 0 (−2)(5) = (3)(5) = 15.
6. Suppose that lim f (x) = 3, lim g(x) = 0, and both f and g are continuous at
x−→2
x−→2
x = 2. Let F (x) := f (x) − g(x).
(a) Find F (2).
Solution. Since f and g are continuous at x = 2, we have that
f (2) = lim f (x) = 3 and g(2) = lim g(x) = 0, hence
x−→2
x−→2
F (2) = f (2) − g(2) = 3 − 0 = 3.
(b) If f were discontinuous at x = 2, can you still find F (2)? Explain.
Solution. No, we could not find the value of F (2) since we wouldn’t even
know whether f (2) is defined.
(c) If f were discontinuous at x = 2, can you find lim F (x)? Explain.
x−→2
Solution. Yes, we can.
lim F (x) = lim (f (x) − g(x)) = lim f (x) − lim g(x) = 3 − 0 = 3.
x−→2
x−→2
x−→2
x−→2
(d) Assume that h is continuous at x = 2. If lim (F + h)(x) = 4, find h(2).
x−→2
Solution. From (a) we know that F (2) = 3. Since both F and h are
continuous at x = 2, we have that
4 = lim (F + h)(x) = lim F (x) + lim h(x) = F (2) + h(2) = 3 + h(2) which
x−→2
x−→2
x−→2
implies that h(2) = 4 − 3 = 1.
7. If f (x) = ex g(x), where g(0) = 2 and g 0 (0) = 5, find f 0 (0).
Solution. f (x) = ex g(x) =⇒ f 0 (x) = (ex )0 g(x) + ex g 0 (x) = ex g(x) + ex g 0 (x).
Hence, f 0 (0) = e0 g(0) + e0 g 0 (0) = g(0) + g 0 (0) = 2 + 5 = 7.
8. Find the point of the curve
y 2 = 2x3
at which the tangent line is perpendicular to the line
4x − 3y + 2 = 0.
Solution. The line 4x − 3y + 2 = 0 can be rewritten as y = (4/3)x + 2/3. Thus it
has slope 4/3. We need the point (a, b) at which the slope m = −3/4. The slope is
y 0 . We use implicit differentiation to obtain y 0 :
(y 2 )0 = (2x3 )0 ⇐⇒ 2yy 0 = 6x2 ⇐⇒ y 0 = 3x2 /y. Thus we need
3x2 /y = −3/4 =⇒ y = −4x2 . Substituting in y 2 = 2x3 we obtain
16x4 = 2x3 =⇒ x = 0 or x = 1/8. If x = 0, then y 0 = 0 6= −3/4 so x = 0 is not
part of the solution. If x = 1/8, then y = −1/16 and with these values we obtain
y 0 = 3(1/8)2 /(−1/16) = −3/4, as required. So the point is (1/8, −1/16).