So many values are already given

Calculus Level 3

Let f(x)f(x) be a non-constant thrice differentiable function defined on real numbers such that f(x)=f(6x)f(x)=f(6-x) and f(0)=0=f(2)=f(5)f'(0)=0=f'(2)=f'(5). Find the minimum number of values of p[0,6]p \in [0,6] which satisfy the equation (f(p))2+f(p)f(p)=0(f''(p))^2+f'(p)f'''(p)=0 Details and Assumptions:

  • f(p)=(df(x)dx)x=pf'(p)=\left( \frac{df(x)}{dx} \right)_{x=p}

  • f(p)=(d2f(x)dx2)x=pf''(p)=\left( \frac{d^2f(x)}{dx^2} \right)_{x=p}

  • f(p)=(d3f(x)dx3)x=pf'''(p)=\left( \frac{d^3f(x)}{dx^3} \right)_{x=p}


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