Question

Function $f\left(x\right)$ is defined as $f\left(x\right)=\{\begin{array}{c}\frac{x-1}{{2x}^2-7x+5},x\ne 1\\ -\frac{1}{3},x=1\end{array}$
Is $f\left(x\right)$ differentiable at $x=1$ if yes find $f^{\prime }\left(1\right)$.

Answer 1

We have,

$f\left(x\right)=\{\begin{array}{c}\frac{x-1}{2x^2-7x+5},x\ne 1\\ -\frac{1}{3},x=1\end{array}$

$\underset{x\to 1}{lim}\frac{\left(x-1\right)}{\left(x-1\right)\left(2x-5\right)}=-\frac{1}{3}$

As $f:x$ is continuous.

$f^{\prime }\left(1^{+}\right)={lim}_{x\to 0}\frac{f\left(1+h\right)-f\left(1\right)}{x}=\frac{\frac{h!}{h\left(2h+3\right)}+\frac{1}{3}}{h}$

$\begin{array}{c}{lim}_{x\to 0}\frac{3+2h-3}{3h\left(2h-3\right)}=\frac{2}{3\left(-1\right)}=\frac{-2}{9}\\ f^{\prime }\left(1^{-}\right)={lim}_{x\to 0}\frac{f\left(1\right)-f\left(1-h\right)}{h}={lim}_{h\to 0}\frac{-\frac{1}{3}-\frac{1}{-3-2h}}{h}\\ =\frac{\frac{1}{3}+\frac{1}{3+2h}}{h}=\frac{-3-2h+3}{3\left(3+2h\right)h}={lim}_{x\to 3}\frac{-2}{3\left(3+2h\right)}\end{array}$

$=\frac{-2}{9}$

Therefore $f\left(x\right)$ is differentiable $f^{\prime }\left(1\right)=\frac{-2}{9}$.