Question

If $f\left(x\right)=\{\begin{array}{c}\frac{x-1}{2x^2-7x+5},x\ne 0\\ \frac{-1}{3},x=1\end{array}$, then $f^{\prime }\left(1\right)$ equals-

• A. $-\frac{2}{9}$
• B. $-\frac{1}{16}$
• C. $0$
• D. does not exist

Answer 1

The correct option is A $-\frac{2}{9}$

$f^{\prime }\left(x\right)=\frac{(2x^2-7x+5)-(x-1)(4x-7)}{(2x^2-7x+5{)}^2}$

$\underset{x\to 1}{lim}\frac{-2x^2+4x-2}{(2x^2-7x+5{)}^2}=\underset{x\to 1}{lim}\frac{-4x+4}{2(2x^2-7x+5)(4x-7)}=\underset{x\to 1}{lim}\frac{-4}{2(4x-7{)}^2}=\frac{-2}{9}$