Suppose f(x) is differentiable for all x. Let g(x) = f(x) +f(1-x^2). Prove that g'(c)=0 for a + real number c.

Question

I tried applying the mean value theorem (& rolle's th'm) as well as the intermediate value th'm but it isn't making sense. There are no solutions, but I have a sense that one of the above methods should be used to solve this question (I may be wrong). Please help!

Answer 1

Clearly g is differentiable.

If we can find a and b such that g(a) = g(b), I'm sure you can finish the proof.

But in fact g(0) = f(0) + f(1) = g(1).

There you are!

Answer 2

Use conjugates. lim(x??) [?(x^2 + gx) - ?(x^2 + hx)] = lim(x??) [?(x^2 + gx) - ?(x^2 + hx)] * [?(x^2 + gx) + ?(x^2 + hx)]/[?(x^2 + gx) + ?(x^2 + hx)] = lim(x??) [(x^2 + gx) - (x^2 + hx)] / [?(x^2 + gx) + ?(x^2 + hx)] = lim(x??) (g - h) x / [?(x^2 (a million + g/x)) + ?(x^2 (a million + h/x))] = lim(x??) (g - h) x / {x [?(a million + g/x) + ?(a million + h/x)]}, in view that x > 0 = lim(x??) (g - h) / [?(a million + g/x) + ?(a million + h/x)] = (g - h)/(a million + a million) = (g - h)/2. i'm hoping this helps!