MATH 409–503
November 14, 2013
Test 2: Solutions
Problem 1 (20 pts.) Find min x3x .
x>0
Solution: min x3x = e−3/e .
x>0
3x
The function f (x) = x3x is well defined and positive on (0, ∞). Hence f (x) = elog f (x) = elog x =
e3x log x for all x > 0. Now it follows from the Chain Rule and the Product Rule that f is differentiable
on (0, ∞) and
f 0 (x) = e3x log x (3x log x)0 = x3x (3x)0 log x + 3x(log x)0 = x3x (3 log x + 3) = 3x3x (log x + 1)
for all x > 0. Notice that a function g(x) = log x + 1 is strictly increasing on (0, ∞). Since g(1/e) = 0,
this implies that g(x) > 0 for x > 1/e and g(x) < 0 for 0 < x < 1/e. Consequently, f 0 (x) > 0 for
x > 1/e and f 0 (x) < 0 for 0 < x < 1/e. It follows that the function f is strictly decreasing on the
interval (0, 1/e] and strictly increasing on the interval [1/e, ∞). In particular, f (x) > f (1/e) for each
x > 0, x 6= 1/e. Thus min f (x) = f (1/e) = (1/e)3/e = e−3/e .
x>0
Problem 2 (25 pts.) Consider a function f : R → R given by

 sin x if x 6= 0,
x
f (x) =

1 if x = 0.
(i) Prove that f is differentiable at 0 and find f 0 (0).
(ii) Prove that f is twice differentiable at 0 and find f 00 (0).
Solution: f 0 (0) = 0, f 00 (0) = −1/3.
For any x 6= 0,
f (x) − f (0)
1
=
x−0
x
sin x − x
sin x
−1 =
.
x
x2
Functions h1 (x) = sin x − x and h2 (x) = x2 are infinitely differentiable on R. Since h1 (0) = h2 (0) = 0,
we have lim h1 (x) = lim h2 (x) = 0. Further, h01 (x) = cos x − 1 and h02 (x) = 2x for all x ∈ R. In
x→0
x→0
particular, h01 (0) = h02 (0) = 0 so that lim h01 (x) = lim h02 (x) = 0. Further, h001 (x) = − sin x and
x→0
x→0
h002 (x) = 2 for all x ∈ R. In particular, h001 (0) = 0 so that lim h001 (x) = 0. Finally, using l’Hôpital’s Rule
x→0
(twice) and a limit theorem, we obtain
lim h00 (x)
h1 (x)
h01 (x)
h001 (x)
0
f (x) − f (0)
x→0 1
= lim
= lim 0
= lim 00
=
= = 0.
f (0) = lim
00
x→0
x→0
x→0
x→0
x−0
h2 (x)
h2 (x)
h2 (x)
lim h2 (x)
2
0
x→0
1
Thus f is differentiable at 0. As a consequence, f is continuous at 0, that is,
lim
x→0
sin x
= 1.
x
By the Quotient Rule, the function f is also differentiable at any x 6= 0 and
sin x 0 (sin x)0 · x − sin x · x0
x cos x − sin x
0
=
.
f (x) =
=
2
x
x
x2
Notice that lim (x cos x − sin x) = 0. Using l’Hôpital’s Rule, we obtain
x→0
f 0 (x) − f 0 (0)
x cos x − sin x
(x cos x − sin x)0
= lim
=
lim
x→0
x→0
x→0
x−0
x3
(x3 )0
f 00 (0) = lim
1
sin x
1
−x sin x
= − · lim
=− .
2
x→0
3x
3 x→0 x
3
= lim
Problem 3 (20 pts.)
Prove the following version of the Fundamental Theorem of
Calculus: if a function F is differentiable on [a, b] and the derivative F 0 is integrable on [a, b],
then
Z b
F 0 (t) dt = F (b) − F (a).
a
Consider an arbitrary partition P = {x0 , x1 , . . . , xn } of the interval [a, b], where a = x0 < x1 <
x2 < · · · < xn = b. Let us choose samples tj ∈ [xj−1 , xj ] for the Riemann sum S(F 0 , P, tj ) so that
F (xj )−F (xj−1 ) = F 0 (tj ) (xj −xj−1 ), j = 1, 2, . . . , n (this is possible due to the Mean Value Theorem).
Then
S(F 0 , P, tj ) =
n
X
j=1
F 0 (tj ) (xj −xj−1 ) =
n
X
j=1
(F (xj ) − F (xj−1 )) = F (xn ) − F (x0 ) = F (b) − F (a).
Since the sums S(F 0 , P, tj ) should converge to the integral
Rb
a
F 0 (t) dt as kP k → 0, the theorem follows.
Problem 4 (25 pts.) Find an indefinite integral and evaluate definite integrals:
Z
Z 1/2
Z 1/2 √
2
2
1 − x2 dx.
(i) x sin x dx,
(ii)
dx,
(iii)
1 − x2
0
0
Solution:
Z
2
2
x sin x dx = −x cos x + 2x sin x + 2 cos x + C,
√
Z 1/2 √
3
π
1 − x2 dx =
+
.
12
8
0
2
Z
1/2
0
2
dx = log 3,
1 − x2
To find the indefinite integral of the function y(x) = x2 sin x, we integrate by parts twice:
Z
Z
Z
Z
2
2
0
2
2
x sin x dx = − x (cos x) dx = − x d(cos x) = −x cos x + cos x d(x2 )
2
= −x cos x +
Z
Z
2
2x cos x dx = −x cos x + 2 x d(sin x)
Z
2
= −x cos x + 2x sin x − 2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C.
To integrate a rational function y(x) = 2/(1 − x2 ), we expand it into a sum of simple fractions:
2
2
1
1
=
=
+
.
2
1−x
(1 − x)(1 + x)
1+x 1−x
Then
1/2
1/2
1/2
1
1/2
1/2
dx = log(1 + x)|x=0 − log(1 − x)|x=0
1
−
x
0
0
0
1
3
= log − log 1 − log − log 1 = log 3.
2
2
√
To integrate the function y(x) = 1 − x2 , we use a substitution x = sin t (observe that x changes
from 0 to 1/2 when t changes from 0 to π/6):
Z π/6 p
Z π/6 p
Z 1/2 p
2
2
1 − x dx =
1 − sin t d(sin t) =
1 − sin2 t (sin t)0 dt
Z
2
dx =
1 − x2
Z
1
dx +
1+x
Z
0
0
=
Z
0
π/6 √
0
cos2 t · cos t dt =
=
Z
π/6
cos2 t dt =
Z
0
0
π/6
√
π
sin(2t) π/6
3
t
=
+
+
.
2
4
12
8
t=0
1 + cos(2t)
dt
2
Bonus Problem 5 (15 pts.) Prove that the sequence {yn } is bounded, where
yn = 1 +
1 1
1
+ + · · · + − log n, n = 1, 2, 3, . . .
2 3
n
The function h(x) = 1/x is integrable on every interval J = [a, b] ⊂ (0, ∞). Hence for any partition
P of the interval J the lower Darboux sum L(h, P ) and the upper Darboux sum U (h, P ) satisfy
Z b
h(x) dx ≤ U (h, P ).
L(h, P ) ≤
a
In the case J = [1, n], where n > 1 is an integer, and P = {1, 2, . . . , n} we obtain
Z n
1
1
1
1 1
h(x) dx = log n,
U (h, P ) = 1 + + · · · +
.
L(h, P ) = + + · · · + ,
2 3
n
2
n−1
1
Then the above inequalities imply that 1/n ≤ yn ≤ 1. Thus the sequence {yn } is bounded.
3