How do I find f'(x) of f(x) = |x+3| -1?
Hello,
write f without absolute values:
$\displaystyle f(x) = |x+3| - 1 = \left\{\begin{array}{lr}x+3-1, & x \geq -3 \\ -(x+3)-1, & x<-3 \end{array}\right.$
Now you can calculate the derivative
$\displaystyle f'(x)=\left \{ \begin{array}{lr}1 & x\geq -3 \\-1, & x < -3\end{array} \right.$
Since x has the coefficient 1 in your original equation the derivative could be only 1 or -1. You only have to determine the point where the slope is changing: The zeros of the absolute value.