Question

Suppose $f\left(x\right)=\frac{x-1}{2x^2-7x+5}$ for $x\ne 1$ and $f\left(1\right)=-\frac{1}{3}$, then

• A. $f$ is continuous but not differentiable at $x=1$
• B. $f$ is differentiable $x=1$ and $f^{\prime }\left(1\right)=-\frac{1}{3}$
• C. $f$ is differentiable $x=1$ and $f^{\prime }\left(1\right)=-\frac{2}{9}$
• D. $f$ is discontinuous at $x=\frac{5}{2}$

Answer 1

Answer: (C), (D)

$f$ is discontinuous at $x=\frac{5}{2}$

$f\left(x\right)=\frac{x-1}{2x^2-7x+5}\Rightarrow f\left(x\right)=\frac{x-1}{(x-1)(2x-5)}$
Checking for continuity,
$x=1\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}\frac{x-1}{(x-1)(2x-5)}=-\frac{1}{3}=f\left(1\right)$
So continuous at $x=1$
Now,
$x=\frac{5}{2}\underset{x\to 5/2}{lim}f\left(x\right)=\underset{x\to 5/2}{lim}\frac{x-1}{(x-1)(2x-5)}\to \text{Not defined}$
So discontinuous at $x=\frac{5}{2}$

Now for finding the derivative of the function at $x=1$
$f\left(x\right)=\frac{1}{2x-5}\Rightarrow f^{\prime }\left(x\right)=-\frac{2}{(2x-5{)}^2}\Rightarrow f^{\prime }\left(1\right)=-\frac{2}{9}$