f(x)=x2−3x+9 then the derivative of  f(x) in the interval  (0,5) is

# $f\left(x\right)=\sqrt{{x}^{2}-3x+9}$ then the derivative of  $f\left(x\right)$ in the interval  $\left(0,5\right)$ is

1. A

0

2. B

1

3. C

$-1$

4. D

Does not exist

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### Solution:

$f\left(x\right)=\sqrt{{x}^{2}-3x+9}$

$=\sqrt{{\left(x-3\right)}^{2}}$

$f\left(x\right)=|x-3|$

$f\left(x\right)=\left\{\begin{array}{l}x-3,x\ge 3\\ 3-x,x<3\end{array}$

$⇒{f}^{|}\left(x\right)=\left\{\begin{array}{cc}1,& if\text{\hspace{0.17em}}x\ge 3\\ -1,& if\text{\hspace{0.17em}}x<3\end{array}$

$f\left(x\right)$ is differentiable at all points on the interval $\left(0,5\right)$ except at  $x=3$

$\therefore$ the derivative of $f\left(x\right)$ on the interval $\left(0,5\right)$ does not exist.

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