The function f(x)=|x^{2}-2 x-3| cdot e^{|9 x^ | vedclass

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(C) Given $f(x)=|x^{2}-2 x-3| \cdot e^{|9 x^{2}-12 x+4|}$.We can factorize the expressions inside the absolute values:$x^{2}-2 x-3 = (x-3)(x+1)$$9 x^{

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(C) Given $f(x)=|x^{2}-2 x-3| \cdot e^{|9 x^{2}-12 x+4|}$.We can factorize the expressions inside the absolute values:$x^{2}-2 x-3 = (x-3)(x+1)$$9 x^{2}-12 x+4 = (3 x-2)^{2}$So,$f(x)=|(x-3)(x+1)| \cdot e^{(3 x-2)^{2}}$.Note that $e^{(3 x-2)^{2}}$ is a smooth,differentiable function for all $x \in \mathbb{R}$.The non-differentiability of $f(x)$ depends solely on the term $|(x-3)(x+1)|$.$A$ function of the form $|g(x)|$ is non-differentiable at the roots of $g(x)$ where the sign changes.Here,$g(x) = (x-3)(x+1)$ changes sign at $x=3$ and $x=-1$.At $x=3$ and $x=-1$,the function $|(x-3)(x+1)|$ has sharp corners (cusps).Therefore,$f(x)$ is not differentiable at exactly two points: $x=3$ and $x=-1$.

The function $f(x)=|x^{2}-2 x-3| \cdot e^{|9 x^{2}-12 x+4|}$ is not differentiable at exactly :

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