Solving a Real-Valued Differentiable Function Problem

ludi_srbin
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Alright, here is the problem.

For all real numbers x, f is 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) f'(0)

c) If L1 and L2 are lines tangent to the graph of f at (-p,1) and (p,1) respectively, and if L1 and L2 intersect at point x- and y- coordinates of Q in terms of p.
 
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O yeah I forgot. I got the a) part easily. Others I can't do.
 
For part b), try using that fact that 0 = - 0 and the properties you should have deduced about the derivative of an even function.

For part c), just construct the two tangent lines. You have enough information to do this if you use the symmetry of the function and its derivative.
 
Thanks man. I appreciate it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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