Question

Assume f′(c) = -4. Find f′(-c) if f is an odd function.

I know an odd function is defined by the given:
f(x) = f(-x) or f(x)+f(-x)=0
Does that apply to its derivative as well?

Answer 1

Ah, it says f is an odd function, not f'. I apologize.

Answer 2

Odd functions have f(-x)=-f(x). For example, sin(-x) = -sin(x), so sine is an odd function.
If f(x) = sin(-x), then f'(x) = cos(x) and then f'(-x) = cos(-x) = cos(x), an even function because f'(-x) = f'(x). So it's not true that if f(x) is odd, so is its derivative. But is the derivative of an odd function always even?

What do you think?

Answer 3

Well if you think about y = x^3, the derivative y= 3x^2 at a point like x=2 will be positive. The derivative at x=-2 will also be positive.

Answer 4

eg.
https://www.google.com/search?q=x%5E3&oq=x%5E3&aqs=chrome.0.69i57j69i65l3j0j5.1466j0&sourceid=chrome&ie=UTF-8

The slope at a point x=a is the same as the slope at x=-a. For example, the slope at x=-3 is the same as at x=3.

Answer 5

Proof that derivative of odd function is even:
\[f'(x_0):=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{-f(x)+f(x_0)}{-x+x_0}=\]
\[=\lim_{x\to x_0}\frac{f(-x)-f(-x_0)}{(-x)-(-x_0)}\stackrel{-x\to y}=\lim_{y\to -x_0}\frac{f(y)-f(-x_0)}{y-(-x_0)}=:f'(-x_0)\]
So for our case f(x) is odd, so f'(x) will be even. Then f'(-c)=f'(c)=-4

Answer 6

@ybarrap @agent0smith both of your replies make sense. From how you guys explained it, the derivative of f′(-c) is clearly -4.

Answer 7

yep