Question

If the following function is differentiable at x = 2, then find the value of a and b :
$f\left(x\right)=\{\begin{array}{c}{x}^2.ifx\le 2\\ ax+b,ifx>2\end{array}$

Answer 1

As the function f is differentiable at x = 2, so it is contiuous at x = 2 as well.
$\therefore {lim}_{x\to 2^{-}}f\left(x\right)={lim}_{x\to 2^{+}}f\left(x\right)=f\left(2\right)\Rightarrow {lim}_{x\to 2^{-}}{x}^2={lim}_{x\to 2^{+}}ax+b=(2{)}^2\Rightarrow 4=2a+b...(i)\text{Also, f is differentiable at x = 2}\therefore L{f}^{\prime }(2)=R{f}^{\prime }(2)i.e.,{lim}_{x\to 2^{-}}\frac{f\left(x\right)-f\left(2\right)}{x-2}={lim}_{x\to 2^{+}}\frac{f\left(x\right)-f\left(2\right)}{x-2}\Rightarrow {lim}_{x\to 2^{-}}\frac{x^2-4}{x-2}={lim}_{x\to 2^{+}}\frac{(ax+b)-4}{x-2}\Rightarrow {lim}_{x\to 2^{-}}(x+2)={lim}_{x\to 2^{+}}\frac{(ax+b)-4}{x-2}[by\left(i\right),b=4-2a]\therefore 4={lim}_{x\to 2^{+}}\frac{(ax+4-2a)-4}{x-2}\Rightarrow 4={lim}_{x\to 2^{+}}\frac{(x-2)a}{x-2}={lim}_{x\to 2^{+}}a\Rightarrow a=4\text{Replacing value of a in (i), we get :}b=-4$