Math 3210-3
Exam 3
Solutions
Name:
You may use your dictionary of definitions. Please use a pencil and keep your proofs neat
and organized. Make sure you use complete “sentences,” and remember you need to give
justifications for each step in your proofs. Each problem is worth one point. Write on this
front page which problems you want graded. Notice that you have some choice as to which
problems you will do. You may refer to theorems we proved in class by either stating the
name of the theorem or giving a brief synopsis of the statement of the theorem.
Grade Problems:
Do one of the following:
1. State the Cauchy Mean Value Theorem.
Let f and g be functions that are continuous on [a, b] and differentiable on (a, b). Then there exists at
least one point c ∈ (a, b) such that [f (b) − f (a)]g ′ (c) = [g(b) − g(a)]f ′ (c).
2. State the Fundamental Theorem Of Calculus. (Either of the two.)
The First Fundamental Theorem of Calculus: Let f be integrable on [a, b]. For each x ∈ [a, b] let
Z b
f (t) dt. Then F is uniformly continuous on [a, b]. Furthermore, if f is continuous at
F (x) =
a
c ∈ [a, b] then F is differentiable at c and F ′ (c) = f (c).
The Second Fundamental Theorem of Calculus: If f is differentiable on [a, b] and f ′ is integrable on
Z b
f = f (b) − f (a).
[a, b], then
a
Do one of the following:
3. Prove that the poloynomial function fm (x) = x3 − 3x + m never has two roots in [0, 1], no matter what
m may be.
Proof: Suppose fm (x) has two roots in [0, 1]. Thus there is some x, y ∈ [0, 1] such that fm (x) =
m (x)
′
fm (y) = 0. By the mean value theorem, there is some c ∈ (x, y) such that fm
(c) = fm (y)−f
= 0.
y−x
′
2
But fm (c) = 3c − 3 = 3(c + 1)(c − 1) = 0 which implies that c = 1 or c = −1, a contradiction.
Therefore fm (x) never has two roots in [0, 1].
4. Let f (x) = sin−1 x. What is f ′ (x)? (Hint: At some point it might be helpful to complete the diagram
below. And just writing the formula you might remember from Calculus does not count.)
Proof: Notice that f (x) is the inverse function of g(θ) = sin θ. So by the Inverse Function Theorem,
1
1
1
1
=
=
.
=√
f ′ (x) = ′
−1
g (θ)
cos θ
cos(sin x)
1 − x2
1
x
1-x2
Prove one of the following:
5. (a) If h(x) ≥ 0 for all x ∈ [a, b], then
Z
b
a
h(x) dx ≥ 0.
Proof: Let P be the partition P = {a, b}. Then there exists x1 ∈ [a, b] such that m = h(x1 ) ≤
Rb
h(x) for all x ∈ [a, b]. Now m ≥ 0, so 0 ≤ L(h, P ) = m(b − a) ≤ a h(x) dx.
(b) If f (x) ≥ g(x) for all x ∈ [a, b], then
Z
a
b
f (x) dx ≥
Z
b
g(x) dx.
a
Z
b
h(x) dx =
Proof: Let h(x) = f (x) − g(x). Then h(x) ≥ 0 for all x ∈ [a, b], so by part (a)
a
Z b
Z b
Z b
Z b
Z b
g(x) dx.
f (x) dx ≥
g(x) dx ≥ 0. Therefore
f (x) dx −
(f (x) − g(x)) dx =
a
a
a
a
a
6. If f is Zcontinuous on [−π, π], differentiable
on (−π, π), and f ′ is integrable on [−π, π], then for all
Z π
π
f (x) cos(nx) dx.
f ′ (x) sin(nx) dx = −n
n ∈ N,
−π
−π
Proof:
= sin(nx). Then (f g)′ =R f ′ g + g ′ f by the product rule. Thus f ′ g = (f g)′ − g ′ f =⇒
R ′
RLet g(x)
′
f g = [(f g) − g ′ f ] = f g(b) − f g(a) − g ′ f by the Fundamental Theorem of Calculus. Therefore
Z π
Z π
′
g′ f
(1)
f (x) sin(nx) dx = f (π) sin(nπ) − f (−π) sin(−nπ) −
−π
−π
Z π
n cos(nx)f (x) dx
(2)
= −
−π
Z π
f (x) cos(nx) dx
(3)
= −n
−π
Prove one of the following:
7. Let f be differentiable on an interval I. Then if f ′ (x) < 0 for all x ∈ I, then f is strictly decreasing
on I.
Proof: Let x1 , x2 ∈ I with x1 < x2 . By the Mean Value Theorem there exists c ∈ (x1 , x2 ) such that
f (x2 ) − f (x1 )
f ′ (c) =
. This implies f (x2 )−f (x1 ) = f ′ (c)(x2 −x1 ) < 0 since f ′ (c) < 0 and x2 −x1 > 0.
x2 − x1
Therefore f is strictly decreasing on I.
8. Let f be a monotonic function on [a, b]. Then f is integrable.
Proof: We will assumce f is nondecreasing. The other case follows in the same way. Since f (a) ≤
f (x) ≤ f (b) for all x ∈ [a, b], f is bounded. Given ǫ > 0 there exists k > 0 such that k[f (b) − f (a)] < ǫ.
Let P = {x0 , x1 , . . . , xn } be a partition of [a, b] such that ∆xi P
< k for all i. Since f P
is nondecreasing we
have mi = f (xi−1 ) and Mi = f (xi ). Thus U (f, P )−(f, P ) = (Mi −mi )∆xi < k f (xi )−f (xi−1 ) =
k[f (b) − f (a)] < ǫ. Therefore by Theorem 97, f is integrable.