If f(x)= x^2+6x+9 , then f '(x) is equal to: - Vidyadip

MCQ

If $\text{f}(\text{x})=\sqrt{\text{x}^2+6\text{x}+9},$ then $f\ '(x)$ is equal to:

A
$1$ for $x < -3$
✓
$-1$ for $x < -3$
C
$1$ for all $\text{x}\in\text{R}$
D
None of these.

✓

Answer

Correct option: B.

$-1$ for $x < -3$

We have, $\text{f}(\text{x})=\sqrt{\text{x}^2+6\text{x}+9}$
$=\sqrt{(\text{x}+3)^2}$
$=|\text{x}+3|$
$\text{f}(\text{x})=\begin{cases}\text{x}+ 3,\text{x}\geq-3\\ -\text{x}-3,\text{x}<-3\end{cases}$
$\Rightarrow\text{f}\ '(\text{x})=\begin{cases} 1,\text{x}\geq-3\\ -1,\text{x}<-3\end{cases}$
$\therefore f\ '(x) = -1$ for $x < -3$

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