Question:
Choose the correct alternative in the following:
If $f(x)=\sqrt{x^{2}+6 x+9}$, then $f^{\prime}(x)$ is equal to
A. 1 for $x<-3$
B. $-1$ for $x<-3$
C. 1 for all $x \in R$
D. none of these
Solution:
$f(x)=\sqrt{x^{2}+6 x+9}$
$\Rightarrow f(x)=\sqrt{(x+3)^{2}}$
$\Rightarrow f(x)=|x+3|$
$\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}(\mathrm{x}+3), \mathrm{x}+3 \geq 0 \Leftrightarrow \mathrm{x} \geq-3 \\ -(\mathrm{x}+3), \mathrm{x}+3<0 \Leftrightarrow \mathrm{x}<-3\end{array}\right.$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{c}1, \mathrm{x} \geq-3 \\ -1, \mathrm{x}<-3\end{array}\right.$