Mathematics - Continuity - Differentiability Question with Solution | TestHub

The function $f(x)=\min (x^2,\alpha x+\beta )$ is differentiable everywhere if and only if the line $y=\alpha x+\beta$ is tangent to the parabola $y=x^2$. This condition implies that the discriminant of $x^2-\alpha x-\beta =0$ is zero, so $(-\alpha {)}^2-4(1)(-\beta )=0⟹{\alpha }^2+4\beta =0⟹\beta =-{\alpha }^2/4$. Thus, $f(x)=\min (x^2,\alpha x-{\alpha }^2/4)$. For $g(x)=|f(x)|$ to be differentiable everywhere, $f(x)$ must not change sign (i.e., $f(x)\ge 0$ for all $x$ or $f(x)\le 0$ for all $x$), or if $f(x_0)=0$, then $f^{\prime }(x_0)=0$. The function $f(x)$ is $x^2$ for $x\le \alpha /2$ and $\alpha x-{\alpha }^2/4$ for $x>\alpha /2$. If $\alpha >0$, then $f(x)\ge 0$ for all $x$, because $x^2\ge 0$ and for $x>\alpha /2$, $\alpha x-{\alpha }^2/4>\alpha (\alpha /2)-{\alpha }^2/4={\alpha }^2/4>0$. In this case, $|f(x)|=f(x)$, which is differentiable everywhere. If $\alpha <0$, let $\alpha =-k$ for $k>0$. Then $f(x)=\min (x^2,-kx-k^2/4)$. This function becomes zero at $x=-k/4$ (since $-k/4>-k/2$, this point is in the linear part). At $x=-k/4$, $f^{\prime }(-k/4)=-k=\alpha \ne 0$. Since $f(-k/4)=0$ but $f^{\prime }(-k/4)\ne 0$, $g(x)=|f(x)|$ is not differentiable at $x=-k/4$. Therefore, for $g(x)$ to be differentiable everywhere, we must have $\alpha >0$.