Question

Can I get help solving this graph? I figured out h'(2) does not exist, but am having difficulty finding h'(1) and h'(3).

Answer 1

F(x) is a piecewise function defined as:
f(x) = (3/2)x when 0 <= x <= 2
or
f(x) = -(3/2)x+6 when 2 < x <= 4

g(x) is defined as: 
g(x) = -(1/4)x+1 when 0 <= x <= 4

Based on the graph or through the equations we can say:
f(1) = (3/2)*1 = 1.5
g(1) = -(1/4)*1+1 = 0.75
f(3) = (3/2)*3 = 4.5
g(3) = -(1/4)*3+1 = 0.25

And the derivative values are:
f ' (1) = 3/2 = 1.5
f ' (3) = -3/2 = -1.5
g ' (1) = -1/4 = -0.25
g ' (3) = -1/4 = -0.25
which are the slopes of each line

So...
h(x) = f(x)*g(x)
h ' (x) = f ' (x)*g(x) + f(x)*g ' (x) ... product rule
h ' (1) = f ' (1)*g(1) + f(1)*g ' (1)
h ' (1) = 1.5*0.75 + 1.5*(-0.25)
h ' (1) = 0.75

and,
h(x) = f(x)*g(x)
h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)
h ' (3) = f ' (3)*g(3) + f(3)*g ' (3)
h ' (3) = -1.5*0.25 + 4.5*(-0.25)
h ' (3) = -1.5