Math 201
Exam 1 Key
Section 2+3
Fall 2012
1. Suppose that lim f (x) = 3, lim g(x) = 0, and both f and
x−→2
x−→2
g are continuous at 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,
x−→2
x−→2
hence 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)?
x−→2
Explain.
Solution. Yes, we can. lim F (x) = lim (f (x) − g(x)) =
x−→2
x−→2
lim f (x) − lim g(x) = 3 − 0 = 3.
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) =
x−→2
x−→2
x−→2
F (2) + h(2) = 3 + h(2) which implies that
h(2) = 4 − 3 = 1.
2. Let f be a function that is differentiable at x = 2. Assume
that the tangent line L to the graph of f at x = 2 passes
through the points (2, 1) and (3, 3). If you can, answer the
following. If you can’t, explain why.
(a) Find f (2).
Solution. Since L passes through (2, 1) and it is tangent
to the graph of f at x = 2, we conclude that (2, 1)
belongs to the graph of f , hence f (2) = 1.
(b) Find an equation for the tangent line L.
−y0
3−1
Solution. We need the slope: m = xy11−x
=
3−2 = 2. We
0
also need a point that can be either (2, 1) or (3, 3).
Choosing (2, 1), we obtain y − y0 = m(x − x0) or
y − 1 = 2(x − 2). Choosing (3, 3), we obtain
y − y0 = m(x − x0) or y − 3 = 2(x − 3).
(c) Find f 0(2).
Solution. We know that f 0(2) can be interpreted as the
slope of the tangent line to the graph of f at x = 2, i.e.,
L. Since the slope of L is 2, we conclude f 0(2) = 2.
(d) Find f 0(3).
Solution. We cannot answer this one. We don’t even
know if f (3) is defined.
(e) Find an equation for the normal line to the graph of f at
x = 2.
Solution. If mn denotes the slope of the normal line, then
it satisfies m · mn = −1 =⇒ mn = −1/2. The normal
line must pass through (2, 1), thus we obtain
y − y0 = mn(x − x0) or y − 1 = (−1/2)(x − 2).
(f) Is f continuous at x = 2? Why?
Solution. Yes; since f is differentiable at x = 2, it must
be continuous at x = 2.
(g) Find lim f (x).
x−→2
Solution. By the above lim f (x) = f (2) = 1.
x−→2
(h) Is f continuous at x = 3? Why?
Solution. We cannot answer this one. We don’t even
know if f (3) is defined.
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
-5
-5
0
5
(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. Let
f (x) :=
x2 sin x1 if x 6= 0
0
if x = 0
(a) Find functions g(x) and h(x) so that
g(x) ≤ f (x) ≤ h(x),
for −1 < x < 1, with lim g(x) = 0 = lim h(x).
x−→0
2
x−→0
2
Solution. Take g(x) = −x and h(x) = x .
(b) Can you use the information in (a) to find lim f (x)?
x−→0
How?
Solution. Since g(x) ≤ f (x) ≤ h(x) and
lim g(x) = 0 = lim h(x), the Squeeze Theorem implies
x−→0
x−→0
that lim f (x) = 0.
x−→0
(c) Is f continuous at x = 0? Why?
Solution. Yes, because lim f (x) = f (0).
x−→0
6. Find the following and show your work:
x2 − 4x + 3
.
(a) lim
x−→3
x−3
x2 − 4x + 3
with x = 3 is an
Solution. Evaluating lim
x−→3
x−3
indetermined form. But when x −→ 3, then x − 3 is
close to zero, but is not zero, hence
x2 − 4x + 3
(x − 1)(x − 3)
lim
= lim
x−→3
x−→3
x−3
x−3
= lim (x − 1) = 2.
x−→3
(b) Find
√
1+h−1
.
h−→0
h
Solution. Substituting x = 0 we obtain:
√
√
1+h−1
1+0−1
lim
=
h−→0
h
0
1 − (1)
=
0
0
=
0
= I. F
√
1+h−1
Thus we proceed to multiply
by
h
√
1+h+1
1=√
to obtain:
1+h+1
lim
√
lim
h−→0
1+h−1
=
h
=
=
=
=
=
=
√
√
1+h−1
1+h+1
√
lim
h−→0
h
√ 1+h+1
√
( 1 + h − 1)( 1 + h + 1)
√
lim
h−→0
h( 1 + h + 1)
(1 + h) − 1
lim √
h−→0 h( 1 + h + 1)
h
lim √
h−→0 h( 1 + h + 1)
1
lim √
h−→0
1+h+1
1
√
1+0+1
1
2
sin(2x)
(c) lim
.
x−→0
2
Solution.
sin(2x)
sin(2(0))
=
x−→0
2
2
sin(0)
=
2
0
=
2
= 0.
lim