Question

For what choice of a and b is the function $f\left(x\right)=\left\{\begin{array}{ll}{x}^2,& x\le c\\ ax+b,& x>c\end{array}\right.$ is differentiable at x = c

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{l}{x}^2,x\le c\\ ax+b,x>c\end{array}\right.$

It is given that the function is differentiable at x=c.

Every differentiable function is continuous. Therefore, it is continuous at x = c.

Then,

$$\begin{gathered} \underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to c^{+}}{\lim }f\left(x\right)=f\left(c\right) \\ \Rightarrow \underset{x\to c}{\lim }{x}^2=\underset{x\to c}{\lim }ax+b=c^2 \\ \Rightarrow c^2=ac+b....\left(\mathrm{i}\right) \end{gathered}$$

Now, f(x) is differentiable at x = c.

(LHD at x=c) = (RHD at x=c)

$\Rightarrow \underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$

​$\Rightarrow \underset{x\to c}{\lim }\frac{x^2-c^2}{x-c}=\underset{x\to c}{\lim }\frac{ax+b-c^2}{x-c}$
$\Rightarrow \underset{x\to c}{\lim }(x+c)=\underset{x\to c}{\lim }\frac{ax+b-ac-b}{x-c}$ [Using (i)]

$$\begin{gathered} \Rightarrow \underset{x\to c}{\lim }(x+c)=\underset{x\to c}{\lim }a \\ \Rightarrow 2c=a \end{gathered}$$
From (i), we have

$$\begin{gathered} c^2=ac+b \\ \Rightarrow c^2=2c^2+b \\ \Rightarrow b=-c^2 \end{gathered}$$

Hence, when $a=2c$ and $b=-c^2$, the given function is differentiable at x=c.