Proving Strictly Increasing Derivative with Continuous Function at a Point

[H2Pendragon]

Homework Statement

Suppose f is differentiable on J, c is in J⁰ and f'(c) > 0. Show that if f' is continuous at c, then f is strictly increasing on some neighborhood of c

Homework Equations

Strictly increasing: If x < y then f(x) < f(y)
Continuous: For all epsilon > 0 there exists a delta > 0 such that x in D union B(a;delta) implies that |f(x) - f(a)| < epsilon

The Attempt at a Solution

I don't have any attempts to write down here. I'm mainly looking for a push in the right direction. I've been staring at the definitions and just can't see the easiest way to link them.

 

[HallsofIvy]

Homework Statement

Suppose f is differentiable on J, c is in J⁰ and f'(c) > 0. Show that if f' is continuous at c, then f is strictly increasing on some neighborhood of c

Homework Equations

Strictly increasing: If x < y then f(x) < f(y)
Continuous: For all epsilon > 0 there exists a delta > 0 such that x in D union B(a;delta) implies that |f(x) - f(a)| < epsilon

Yes, and now take a= c and epsilon= f(c)/2 to show that f'(x)> 0 in (c-epsilon, c+epsilon).

The Attempt at a Solution

I don't have any attempts to write down here. I'm mainly looking for a push in the right direction. I've been staring at the definitions and just can't see the easiest way to link them.

 

[H2Pendragon]

Yes, and now take a= c and epsilon= f(c)/2 to show that f'(x)> 0 in (c-epsilon, c+epsilon).

Thanks for the reply. How does taking epsilon = f(c)/2 get me that f'(x) > 0?

I plugged it in and I end up getting that f'(c) - f(c)/2 < f(x) < f'(c) + f(c)/2

I understand, though, that finding f'(x) > 0 solves it. I'm just curious about this middle step. I assume I have to show that f'(c) = f(c)/2 ??

 

[H2Pendragon]

Did you mean to let epsilon = f'(c)/2?

Because that would solve it.

It would then be that f'(c) - f'(c)/2 < f'(x) => f(c)/2 < f'(x). Which means that f'(x) is strictly greater than 0. Thus f(x) is strictly increasing.

Is this right?