Mathematics - Continuity - Differentiability Question with Solution | TestHub

The function $f(x)$ is defined piecewise based on where $|x|$ or $x^2-1$ is greater. The points of intersection of $y=|x|$ and $y=x^2-1$ are $x_1=\frac{1+\sqrt{5}}{2}$ and $x_2=\frac{-1-\sqrt{5}}{2}$. Thus, $f(x)=x^2-1$ for $x\in (-\infty ,x_2]\cup [x_1,\infty )$, and $f(x)=|x|$ for $x\in (x_2,x_1)$. The potential points of non-differentiability are $x_1,x_2$, and $x=0$ (where $|x|$ changes definition). Evaluating left and right derivatives at these points shows $f(x)$ is not differentiable at $x_1,x_2,0$. For instance, at $x=0$, $f^{\prime }(0^{-})=-1$ and $f^{\prime }(0^{+})=1$, so option B is incorrect. For $x\in (0,x_1)$, $f(x)=x$, so $f^{\prime }(x)=1$, making option C correct. Since there are exactly three points of non-differentiability ($x_1,x_2,0$), option A is correct. Option D is incorrect as $x_1$ is a point of non-differentiability.