Question

Show that $f\left(x\right)=\left\{\begin{array}{l}12x-13,\mathrm{if}x\le 3\\ 2x^2+5,\mathrm{if}x>3\end{array}\right.$ is differentiable at x = 3. Also, find f'(3).

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{l}12x-13,x\le 3.\\ 2x^2+5,x>3.\end{array}\right.$

We have to show that the given function is differentiable at x = 3.

We have,

(LHD at x=3) = $\underset{x\to 3^{-}}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-3}$

$$\begin{gathered} =\underset{x\to 3}{\lim }\frac{12x-13-23}{x-3} \\ =\underset{x\to 3}{\lim }\frac{12x-36}{x-3} \\ =\underset{x\to 3}{\lim }\frac{12(x-3)}{x-3} \\ =\underset{x\to 3}{\lim }12 \\ =12 \end{gathered}$$

(RHD at x = 3) = $\underset{x\to 3^{+}}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-3}$

$$\begin{gathered} =\underset{x\to 3}{\lim }\frac{2x^2+5-23}{x-3} \\ =\underset{x\to 3}{\lim }\frac{2x^2-18}{x-3} \\ =\underset{x\to 3}{\lim }\frac{2(x^2-9)}{x-3} \\ =\underset{x\to 3}{\lim }2(x+3) \\ =2\times 6 \\ =12 \end{gathered}$$

Thus, (LHD at x=3) = (RHD at x=3) = 12.

So, $f\left(x\right)$ is differentiable at x=3 and $f\text{'}\left(3\right)=12.$