Question

100 POINTS

The graph of f ′ (x), the derivative of f(x), is continuous for all x and consists of five line segments as shown below. Given f (0) = 6, find the absolute maximum value of f (x) over the interval [0, 5].

graph of line segments increasing from x equals negative 4 to x equals negative 3, decreasing from x equals negative 3 to x equals 0, increasing from x equals 0 to x equals 3, constant from x equals 2 to x equals 4 and decreases from x equals 4 to x equals 5. x intercepts at x equals negative 4, x equals 0, x equals 5

2
8
11
13

Answer 1

The graph of f'(x) is given, so if you find the area under the graph, that will be the value of f(x). Calculate the area of the shape under the graph from x values 0 to 4. This shape should look like a trapezoid. The area you find is actually the change in f(x) from 0 to 4. You will need to add that area to the initial value the problem gave you: f(0)=7.

Area of a trapezoid = h(b2+b2)/2
Height of the trapezoid = 2
Base 1= 2
Base 2 = 4
Area of trapezoid from [0,4] = 6

Initial value: f(0)=7

Absolute Maximum F(x): 6 + 7 = 13