Question

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

• A. $1forx>-3$
• B. $-1forx<-3$
• C. $1forallx\in R$
• D. $0forx=-3$

Answer 1

Answer: (A), (B)

$-1forx<-3$

$f\left(x\right)=\sqrt{x^2+6x+9}=\left(x+3\right|\mathrm{Function}\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=-3.f\left(x\right)=(\begin{array}{c}-(x+3),\mathrm{if}x<-3\\ \left(x+3\right),\mathrm{if}x>-3\end{array}\Rightarrow f\text{'}(x)=(\begin{array}{c}-1,\mathrm{if}x<-3\\ 1,\mathrm{if}x>-3\end{array}$