Mathematics - Continuity - Differentiability Question with Solution | TestHub

From $f(x+y)=f(x)+f(y)+2xy$, setting $x=y=0$ gives $f(0)=0$. The derivative $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}={\lim }_{h\to 0}\frac{f(x)+f(h)+2xh-f(x)}{h}={\lim }_{h\to 0}\left(\frac{f(h)}{h}+2x\right)$. Since $f(0)=0$, ${\lim }_{h\to 0}\frac{f(h)}{h}=f^{\prime }(0)$. Given $f^{\prime }(0)=3$. So $f^{\prime }(x)=3+2x$. Integrating, $f(x)=3x+x^2+C$. Since $f(0)=0$, $C=0$. Thus $f(x)=x^2+3x=x(x+3)$.

The function $g(x)=|f(x)|=|x(x+3)|$ is not differentiable at points where $f(x)=0$ and $f^{\prime }(x)\ne 0$. $f(x)=0$ implies $x(x+3)=0$, so $x=0$ or $x=-3$. At $x=0$, $f^{\prime }(0)=3\ne 0$. At $x=-3$, $f^{\prime }(-3)=3+2(-3)=-3\ne 0$. Therefore, $g(x)$ is not differentiable at $x=0$ and $x=-3$. There are 2 such points.