Mathematics - Continuity - Differentiability Question with Solution | TestHub

The function is $f(x)=(x^2-9)|x-3|=(x-3)(x+3)|x-3|$. We can rewrite this as $f(x)=\{\begin{array}{ll}(x-3{)}^2(x+3)& ,x\ge 3\\ -(x-3{)}^2(x+3)& ,x<3\end{array}$. For $x\ne 3$, the function is a polynomial and thus differentiable. We only need to check differentiability at $x=3$. The left-hand derivative at $x=3$ is $f^{\prime }(3^{-})={\lim }_{h\to 0^{-}}\frac{-(3+h-3{)}^2(3+h+3)-0}{h}={\lim }_{h\to 0^{-}}\frac{-h^2(h+6)}{h}={\lim }_{h\to 0^{-}}-h(h+6)=0$. The right-hand derivative at $x=3$ is $f^{\prime }(3^{+})={\lim }_{h\to 0^{+}}\frac{(3+h-3{)}^2(3+h+3)-0}{h}={\lim }_{h\to 0^{+}}\frac{h^2(h+6)}{h}={\lim }_{h\to 0^{+}}h(h+6)=0$. Since $f^{\prime }(3^{-})=f^{\prime }(3^{+})=0$, the function is differentiable at $x=3$. Therefore, there are no points where $f(x)$ is not differentiable. The number of such points is 0.