- #1

utkarshakash

Gold Member

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## Homework Statement

Let f(x) be a non-constant twice differentiable function defined on R such that f(x)=f(1-x) and f'(1/4) =0 then what is the minimum number of real roots of the equation (f"(x))^2+f'(x)f'''(x)=0.

## The Attempt at a Solution

f'(x)=-f'(1-x)

f"(x)=f"(1-x)

f'''(x)=-f'''(1-x)

putting x = 1/4 in the first eqn gives f'(3/4)=0

putting x=1/4 in the given equation (f"(x))^2+f'(x)f'''(x)=0 gives f"(1/4)=0. ∴f"(3/4)=0.

So I have got a total of 2 roots. But there are some more which I can't find.