Question

Let a > 0 and f be continuous on [-a, a]. Suppose that f'(x) exists and...

Let a > 0 and f be continuous on [-a, a]. Suppose that f'(x) exists and f'(x)<= 1 for all x2 ㅌ (-a, a). If f(a) = a and f(-a) =-a. Show that f(0) = 0.

Hint: Consider the two cases f(0) < 0 and f(0) > 0. Use mean value theorem to prove that these are impossible cases.

Math   Calculus   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Please UPVOTE if this answer helps you understand better.

Solution:-

Please UPVOTE if this answer helps you understand better.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Suppose f is continuous for x is greater than or equal to 0, f'(x) exists for...
Suppose f is continuous for x is greater than or equal to 0, f'(x) exists for x greater than 0, f(0)=0, f' is monotonically increasing. For x greater than 0, put g(x) = f(x)/x and prove that g is monotonically increasing.
1.- Prove the intermediate value theorem: let (X, τ) be a connected topological space, f: X...
1.- Prove the intermediate value theorem: let (X, τ) be a connected topological space, f: X - → Y a continuous transformation and x1, x2 ∈ X with a1 = f (x1), a2 = f (x2) ( a1 different a2). Then for all c∈ (a1, a2) there is x∈ such that f (x) = c. 2.- Let f: X - → Y be a continuous and suprajective transformation. Show that if X is connected, then Y too.
6. Let a < b and let f : [a, b] → R be continuous. (a)...
6. Let a < b and let f : [a, b] → R be continuous. (a) Prove that if there exists an x0 ∈ [a, b] for which f(x0) 6= 0, then Z b a |f(x)|dxL > 0. (b) Use (a) to conclude that if Z b a |f(x)|dx = 0, then f(x) := 0 for all x ∈ [a, b].
Let f be a function for which the first derivative is f ' (x) = 2x...
Let f be a function for which the first derivative is f ' (x) = 2x 2 - 5 / x2 for x > 0, f(1) = 7 and f(5) = 11. Show work for all question. a). Show that f satisfies the hypotheses of the Mean Value Theorem on [1, 5] b)Find the value(s) of c on (1, 5) that satisfyies the conclusion of the Mean Value Theorem.
Let f be monotone increasing and absolutely continuous on [0, 1]. Let E be a subset...
Let f be monotone increasing and absolutely continuous on [0, 1]. Let E be a subset of [0, 1] with m∗(E) = 0. Show that m∗(f(E)) = 0. Hint: cover E with countably many intervals of small total length and consider what f does to those intervals. Use Vitali Covering Argument
f(x) ≥ 0 for all x ∈ (0, 1) and its third derivative f ^3(3)(x) exists...
f(x) ≥ 0 for all x ∈ (0, 1) and its third derivative f ^3(3)(x) exists for all x ∈ (0, 1). If f(x) = 0 for two different values of x in (0, 1), prove that there exists a c in (0, 1) such that f ^(3)(c) = 0.
Let f : [1, 2] → [1, 2] be a continuous function. Prove that there exists...
Let f : [1, 2] → [1, 2] be a continuous function. Prove that there exists a point c ∈ [1, 2] such that f(c) = c.
Let f be a continuous function on the real line. Suppose f is uniformly continuous on...
Let f be a continuous function on the real line. Suppose f is uniformly continuous on the set of all rationals. Prove that f is uniformly continuous on the real line.
Let f be continuous on [ 0 , ∞ ) and differentiable on ( 0 ,...
Let f be continuous on [ 0 , ∞ ) and differentiable on ( 0 , ∞ ) . If f ( 0 ) = 0 and | f ′ ( x ) | ≤ | f ( x ) | for all x > 0 , then f ( x ) = 0 for all x ≥ 0 .
Let a < b, a, b, ∈ R, and let f : [a, b] → R...
Let a < b, a, b, ∈ R, and let f : [a, b] → R be continuous such that f is twice differentiable on (a, b), meaning f is differentiable on (a, b), and f' is also differentiable on (a, b). Suppose further that there exists c ∈ (a, b) such that f(a) > f(c) and f(c) < f(b). prove that there exists x ∈ (a, b) such that f'(x)=0. then prove there exists z ∈ (a, b) such...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT