AP Calculus BC Test Prep: Solving f(-x)=f(x) Problem | Ryan

[Ryush806]

To review for the AP Calculus BC test coming up in May, my teacher has been giving us problems from past AP tests to help us review. However, I have absolutely no idea how to do on of the problems that I was assigned:

For all real numbers x, f is a differentiable function such that f(-x)=f(x). Let f(p)=1 and f'(p)=5 for some p>0.
a) Find f'(-p).
b) Find f'(0).
c) If line 1 and line 2 are lines tangent to the graph of f at (-p,1) and (p,1), respectively, and if line 1 and line 2 intersect at point Q, find the x- and y-coordinates of Q in terms of p.

I'm sure this problem is not incredibly hard but I'm very much confused. Please help me get started on it.

Ryan

 

[NateTG]

All of these are based on the notion that f is symetric about x=0.

 

[matt grime]

f is symmetric in the sense that f is even - reflective symmetry in the y axis.

take the derivative on both sides of f(x)=f(-x) what do you get?

if you've done that properly what you've got defines another form of symmetry. f' is an odd function. odd functions have rotational symmetry about the origin.

you shoudl be able to find the answers now. remember z=-z when and only when z=0

 

[Ryush806]

Originally posted by matt grime
take the derivative on both sides of f(x)=f(-x) what do you get?

Thank you for your help so far. I'm confused about taking the derivative of f(x)=f(-x). I'm not the best at this theory stuff.

Ryan

 

[matt grime]

f(x)=f(-x)

so d/dx(f(x)) = d/dx(f(-x))

use the chain rule on the rhs if you need to to get

f'(x) = -f'(-x)

or better I reckon -f'(x) = f'(-x)

so f'(-p) = -f'(p)

and -f'(0) = f'(-0) = f'(0)

so it must be that f'(0)=0