Find the values of a and b so that the following function is differentiable for all values of $\displaystyle x$ : $\displaystyle \mathrm{f}(x)= \begin{cases}\mathrm{a} x+\mathrm{b}, & x>-1 \\ \mathrm{~b} x^{2}-3, & x \leq-1\end{cases}$

Since Differentiability ⇒ Continuity $\displaystyle \mathrm{f}(x)$ is continuous at $\displaystyle x=-1 \Rightarrow \lim _{x \rightarrow-^{-}}\left(\mathrm{b} x^{2}-3\right)=\lim _{x \rightarrow-1^{+}}(\mathrm{a} x+\mathrm{b})$\Rightarrow \mathrm{b}-3=-\mathrm{a}+\mathrm{b}, \therefore \mathrm{a}=3$\displaystyle \mathrm{f}(x)$ is differentiable at $\displaystyle x=-1$\begin{aligned} & \Rightarrow \lim _{x \rightarrow-1^{-}} \frac{\left(\mathrm{b} x^{2}-3\right)-(\mathrm{b}-3)}{x+1}=\lim _{x \rightarrow-1^{+}} \frac{(\mathrm{a} x+\mathrm{b})-(\mathrm{b}-3)}{x+1} & \Rightarrow \lim _{x \rightarrow-1^{-}} \mathrm{b}(x-1)=\lim _{x \rightarrow-1^{+}} \frac{3(x+1)}{x+1}, \therefore \mathrm{b}=-\frac{3}{2} \end{aligned}Alternately,L H D=\lim _{h \rightarrow 0} \frac{\mathrm{f}(-1)-\mathrm{f}(-1-h)}{h}=\lim _{h \rightarrow 0} \frac{(\mathrm{b}-3)-\left(\mathrm{b}(-1-h)^{2}-3\right)}{h}=\lim _{h \rightarrow 0}(-\mathrm{b} h-2 \mathrm{b})=-2 \mathrm{b}$\displaystyle \mathbf{R H D}=\lim _{h \rightarrow 0} \frac{\mathrm{f}(-1)-\mathrm{f}(-1-h)}{h}=\lim _{h \rightarrow 0} \frac{[\mathrm{a}(-1+h)+\mathrm{b}]-(\mathrm{b}-3)}{h}=\lim _{h \rightarrow 0} \frac{(-\mathrm{a}+3)+\mathrm{a} h}{h}=\mathrm{a}$\text { and }-\mathrm{a}+3=0 \Rightarrow \mathrm{a}=3,-2 \mathrm{b}=\mathrm{a} \Rightarrow \mathrm{b}=-3 / 2 $$