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Q. Let $y=f(x)$ be a twice differentiable function such that $f(1)=2 ; f(2)=-1 ; f(3)=5 ; f(4)=-$ 3 and $f(5)=1$. Find the minimum number of real roots of the equation $f^{\prime}(x) \cdot f^{\prime \prime}(x)=0$.

Continuity and Differentiability

Solution:

$f (x) = 0$ has atleast 4 real roots (IMVT)
$f '(x) = 0$ has atleast 3 real roots (Rolle's)
$f "(x) = 0$ has atleast 2 real roots (Rolle's)