62.8k views
4 votes
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).

Can I get help solving this graph? I figured out h'(2) does not exist, but am having-example-1

1 Answer

3 votes
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
User Hillin
by
7.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories