1. g”(4) is the slope of f(x) at x=4.
Part (a) 1 2
g”(4) is the slope of f(x) at x=4.
g”(4) =
2 – (-2)
3 – 5
g(4) = f(t) dt 4
= 3
g”(4) = -2
g’(4) = f(4)
= 0

2. Therefore, g(x) has a relative minimum at x=1.
Part (b)
Remember that f(x) is the derivative of g(x). At x=1, the derivative changes signs from negative to positive.
This means that the original function g(x) goes from falling to rising there.
Therefore, g(x) has a relative minimum at x=1.

3. Part (c) 1 2
g(10) = f(t) dt 10
= 4
(10,-2)
2 – (-2)
3 – 5

4. Tangent line: (y-44) = m(x-108) 2
Part (c)
g(105) = f(t) dt
105
5 units wide 2 2 1 1 1 1 1 1 1 1
At g(108), the slope is 2. We know this because f(x) is the derivative (a.k.a.—slope) of g(x). At x=108, we’d be right here, giving us a slope of 2.
(10,-2)
Every 5 units along the x-axis increases the value of the integral by 2, so g(105) would be (21)(2), or 42.
As you can see from the graph, moving up to g(108) tacks on an additional 2 units of area. So g(108) = 44.
2 – (-2)
3 – 5
Tangent line: (y-44) = m(x-108) 2