Math 30G – Calculus – Sample Exam Fall 2012

Math 30G – Calculus – Sample Exam
Fall 2012
The Harvey Mudd College Honor Code applies to this exam. That means
it is a violation of the Honor Code to make this exam accessible in any
format to any person not currently enrolled in this class this semester.
DO NOT RE-DISTRIBUTE THIS SOLUTION FILE
1
20
2
20
3
10
4
10
5
10
Total
70
1
Problem 1 (2 pts each). True or False (with no explanations needed, no partial credit).
(a)
F
If f and g are differentiable, then
d
[f (x)g(x)] = f 0 (x)g 0 (x).
dx
(b)
T
If f and g are differentiable, then
d
[f (g(x))] = f 0 (g(x))g 0 (x).
dx
(c)
T
There exists a function f : R → R such that f is continuous nowhere but |f | is
continuous everywhere.
(d)
F
If a function f is continuous on [0, ∞), then it has an absolute minimum on this
interval.
(e)
F
The slope of the tangent line at a point (a, f (a)) of the graph of a differentiable
function f is equal to f 01(a) .
(f)
F
The sum of the series
∞
X
22k
k=0
Suppose we know that
∞
X
5k
is
1
.
5
ak converges to 0.8. We are given no other information about
k=1
the infinite series. For the remaining parts, determine if the following statements are
• true (i.e., must be true),
• or false (i.e., must be false).
(g)
F
(h)
T
(i)
F
(j)
T
lim ak = 0.8.
k→∞
lim ak = 0.
k→∞
ak+1
> 1.
k→∞ ak
lim
lim Sn = 0.8, where Sn = a1 + a2 + · · · + an .
n→∞
2
Problem 2 (5 pts each, 20 pts total). Short answer.
(a) The function
(
if x < 1
x + b if x ≥ 1
f (x) =
is continuous for every x if b =
x2 −4
x−2
2 .
(b) Assume we have two differentiable functions f and g. We are given that f 0 (x) = g 0 (x),
f (1) = 0, and g(1) = 5. Are f and g related? Can you find an explicit relationship
between f and g?
From a result proven in class, we know that since f 0 (x) = g 0 (x), it follows that
f (x) = g(x) + C for some constant C. We also know that
0 = f (1) = g(1) + C = 5 + C
=⇒
C = −5.
So, the relationship between f and g is f (x) = g(x) − 5 .
√
3
1+h−1
by relating it to the value of a derivative. (This means you should
h→0
h
not use l’Hˆopital’s rule here.)
√
Consider the function f (x) = 3 x. This function is differentiable at x = 1, and the
2
derivative at 1 is f 0 (1) = 13 1− 3 = 13 . By the definition of derivative, we also know that
(c) Find lim
f (1 + h) − f (1)
f (1) = lim
= lim
h→0
h→0
h
0
√
√
√
3
3
1+h− 31
1+h−1
= lim
.
h→0
h
h
Thus, it must be that the value of this limit is
1
since f 0 (1) = 13 .
3
1
= 0 means the following (fill in blanks appropriately):
n→∞ n2
(d) To say that lim
For every
we have
> 0 , there is an index N such that for
all integers n ≥ N
,
1
− 0 < .
n2
(If you just wrote n12 , that is correct also. If you did not include the word
“integers” above, that is fine also.)
To satisfy the above definition, if =
1
,
100
then the smallest integer N can be is
3
11 ..
Problem 3 (10 pts).
(a) Suppose f (x) is a differentiable function. Show that at any x,
f (x + h) − f (x − h)
= f 0 (x).
h→0
2h
lim
Hint: You may find it helpful to add and subtract a value to the numerator in the limit.
Assume f is a differentiable function. Then we have
f (x + h) − f (x) + f (x) − f (x − h) by adding and subtracting
f (x + h) − f (x − h)
= lim
f (x) in numerator
h→0
h→0
2h
2h
f (x + h) − f (x) f (x − h) − f (x)
= lim
−
h→0
2h
2h
f (x + h) − f (x) 1
f (x + (−h)) − f (x) by limit
1
+ lim
= lim
properties
2 h→0
h
2 h→0
−h
1
1
by definition of derivative
= f 0 (x) + f 0 (x)
2
2
= f 0 (x).
lim
(b) Use the limit given in part (a) to find f 0 (x) for the function f (x) = x2 .
For the function f (x) = x2 , we have
f (x + h) − f (x − h)
(x + h)2 − (x − h)2
= lim
h→0
h→0
2h
2h
2
x + 2xh + h2 − (x2 − 2xh + h2 )
= lim
h→0
2h
4xh
= lim
h→0 2h
= lim 2x
lim
h→0
= 2x .
4
Problem 4 (10 pts). Let f be a function having derivatives of all orders for all real numbers.
The third-degree Taylor polynomial for f at the point a = −2 is given by
3
1
T3 (x) = 2 − (x + 2)2 − (x + 2)3 .
8
12
(a) Find f (−2), f 0 (−2), and f 00 (−2).
Using the formula for the third-order Taylor polynomial for f at a = −2, we have
3
X
f (k) (−2)
k=0
k!
3
1
(x + 2)k = T3 (x) = 2 − (x + 2)2 − (x + 2)3 .
8
12
It follows that
f (−2) = 2
f 0 (−2) = 0
f 00 (−2) = −
3
4
(since no (x + 2) term exists)
f 00 (−2)
3
since
=−
.
2!
8
(b) Determine whether f has a local minimum, a local maximum, or neither at the point −2.
Justify your answer.
Since f 0 (−2) = 0 and f 00 (−2) = − 43 < 0, we know that f has a local maximum at
x = −2 by the second derivative test.
(c) Use T3 (x) to find an approximation for f (−1).
For an approximation to f (−1), we have
3
1
3
1
37
f (−1) ≈ T3 (−1) = 2 − (−1 + 2)3 − (−1 + 2)3 = 2 − −
=
.
8
12
8 12
24
(d) Find a best linear approximation to the function f at a = −2.
The best linear approximation to f at the point a = −2 is given by the tangent line to f
at the point −2. Furthermore, this tangent line is contained in the third-order polynomial
and is
T1 (x) = 2 .
5
Problem 5 (10 pts). Use the mean value theorem to prove that
| sin y − sin z| ≤ |y − z| for all y, z ∈ R.
Let [y, z] be any interval of the real line. Since the function f (x) = sin x is continuous and
differentiable over all of R, it is certainly differentiable over (y, z). By the mean value theorem,
there exists some value c ∈ (y, z) such that
f 0 (c) =
f (y) − f (z)
sin y − sin z
=
.
y−z
y−z
Therefore, we have
sin y − sin z = f 0 (c)(y − z)
= (cos c)(y − z) since f 0 (x) = cos x,
which implies that
|sin y − sin z| = |(cos c)(y − z)| = |cos c| |y − z| ≤ |y − z|
since | cos x| ≤ 1 for all x (including x = c). Thus, we have shown that
| sin y − sin z| ≤ |y − z| for all y, z ∈ R.
6