What is the relationship between slope and symmetry in differentiable functions?

Nutz
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:smile: 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 ß1 and ß2 are lines tangent to the graph of f at (-p,1) and (p,1) respectibely, and if ß1 and ß2 intersect at point Q, find the x and y coordinates of Q in terms of p.





SO basically the week we did this stuff I was out with pneumonia for 6 days. Sucks, but know I am stuck with this assignment sheet. I pretty much don't know where to begin on this. Any help would be greatly appreciated.
 
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Nutz said:
:smile: 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 ß1 and ß2 are lines tangent to the graph of f at (-p,1) and (p,1) respectibely, and if ß1 and ß2 intersect at point Q, find the x and y coordinates of Q in terms of p.
SO basically the week we did this stuff I was out with pneumonia for 6 days. Sucks, but know I am stuck with this assignment sheet. I pretty much don't know where to begin on this. Any help would be greatly appreciated.

There was a very similar problem in another forum, and you might be the same person. :confused:

Go check it out, I think it's in General Math. Completely identical.
 
You might consider:

Since f(-x) = f(x), this makes f an even function, i.e., symmetric about the y axis..

symmetry should say something about the value of the slope at equal distances from x=0.
 
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