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Q. If $f$ be a function defined on an open interval $I$. Suppose $c \in I$ be any point. If $f$ has a local maxima or a local minima at $x=c$, then
Statement I$ f^{\prime}(c)=0$.
Statement II $f$ is not differentiable at c.

Application of Derivatives

Solution:

From the previous two solutions, geometrically we can say, if $x=c$ is a point of local maxima of $f$, then the graph of $f$ around ' $c$ ' will be as shown in Fig (a). Note that function $f$ is increasing (i.e., $\left.f^{\prime}(x)>0\right)$ in the interval $(c-h, c)$ and decreasing (i.e., $\left.f^{\prime}(x)<0\right)$ in the interval $(c, c+h)$.
This suggests that $f^{\prime}$ (c) must be zero.
image
Similarly, if ' $c$ ' is a point of local minima of $f$, then the graph of $f$ around ' $c$ ' will be as shown in Fig (b). Here, $f$ is decreasing (i.e., $\left.f^{\prime}(x) < 0\right)$ in the interval $(c-h, c)$ and increasing (i.e., $\left.f^{\prime}(x) > 0\right)$ in the interval $(c, c+h)$. This again suggest that $f^{\prime}(c)$ must be zero.
The above discussion lead us to the following result
Let $f$ be a function defined on an open interval $I$. Suppose $c \in I$ be any point. If $f$ has a local maxima or a local minima at $x=c$, then either $f^{\prime}(c)=0$ or $f$ is not differentiable at $c$.