Question

For the function $f\left(x\right)=x^2-6x+8,2\le x\le 4$, then the value of $x$ for which$f\text{'}\left(x\right)$ vanishes is

• A. $\frac{9}{4}$
• B. $\frac{5}{2}$
• C. $3$
• D. $\frac{7}{2}$

Answer 1

The correct option is C

$3$

Explanation for the correct option:

Step1. Differentiation with respect to $x$.

$f\left(x\right)=x^2-6x+8,2\le x\le 4$

$\Rightarrow$$f\text{'}\left(x\right)=2x-6$

Step 2. Value of $x$ for which$f\text{'}\left(x\right)$ vanishes is

Put $\mathrm{f}\text{'}\left(\mathrm{x}\right)=0$

Then,

$\begin{array}{rcl}2x-6& =& 0\end{array}$

$\Rightarrow$ $\mathrm{x}=3$

Hence, correct option is $\mathbf{\left(}\mathit{C}\mathbf{\right)}\mathbf{.}$