## Calculus: Early Transcendentals 8th Edition

h'(x) = $\frac{|x+2|}{x+2}$ + $\frac{|x-1|}{x-1}$ On the graph below: The black graph is h(x) The blue graph is h'(x)
h(x) = |x-1| + |x+2| = $\Bigg\{ \begin{array}{l} -(x-1) - (x+2) \quad for\thinspace x \leq -2\\ -(x-1)+(x+2) \quad for\thinspace x\thinspace-2\lt x\leq 1\\ +(x-1)+(x+2) \quad for\thinspace x\gt 1 \end{array}$ h(x) = $\Bigg\{ \begin{array}{l} -2x-1 \quad for\thinspace x \leq -2\\ \qquad 3 \qquad for\thinspace x\thinspace-2\lt x\leq 1\\ \thinspace2x+1 \qquad for\thinspace x\gt 1 \end{array}$ Differentiate h(x) to get h'(x) = $\Bigg\{ \begin{array}{l} -2 \quad for\thinspace x \leq -2\\ \space 0 \quad for\thinspace x\thinspace-2\lt x\leq 1 \rightarrow \alpha\\ 2 \quad for\thinspace x\gt 1 \end{array}$ Note that h(x) has an edge at x = -2 and x = 1 Therefore, the derivate does not exist at x = -2 and x = 1 We are asked to give a formula for h'(x) We already know h'(x) for diffrent values of x, now we need to find a formula which satisfies $\alpha$ You can come up with h'(x) = $\frac{|x+2|}{x+2}$ + $\frac{|x-1|}{x-1}$