## anonymous one year ago Recall that a function 'f' is said to be even if f(-x) = f(x) for all x. Suppose we are given an even real-valued function 'f' which is differentiable for all values of x, and suppose furthermore that f(c) = 1, and f'(c) = 5 at some point c > 0. Use the definition of the derivative to find f'(-c). State the value of f'(0), giving reasons for your answer.

1. anonymous

Definition of f' in x=c is: $$f'(c)=\lim_{h\to 0} \dfrac{f(c+h)-f(c)}{h}$$ What you can do now, is replace c with -c in this definition.

2. anonymous

Ah okay, so like: f'(c) = lim f(-c + h) - f(-c) h->0 h

3. anonymous

Yes, $$f'(-c)=\lim_{h \rightarrow 0}\dfrac{f(-c+h)-f(-c)}{h}=\lim_{h \rightarrow 0}\dfrac{f(c-h)-f(c)}{h}=$$ $$\lim_{h \rightarrow 0}\dfrac{f(c--h)-f(c)}{-h}=-\lim_{h \rightarrow 0}\dfrac{f(c+h)-f(c)}{h}=-f'(c)$$