Graphs, Double Prime PLS HELP ASAP !

Question

The graph of f ' (x), the derivative of f of x, is continuous for all x and consists of five line segments as shown below. Given f (0) = 7, find the absolute minimum value of f (x) over the interval [-3, 0].

 (4 points)

Answer Choices:	0
	2.5
	4.5
	11.5

Answer 1

By looking at the graph, on the interval [-3,0]

f'(x) = -x

Therefore f(x), being the integral of f'(x), is -x.x/2 + C.

The constant C can be obtained from the condition "f(0) = 7",

which yields C = 7.

Thus f(x) = -x.x/2 + 7 (on the interval [-3, 0])

On the interval [-3,0] the function f is increasing, therefore

its minimum occurs at x = -3.

f(-3) = 11.5

Answer 2

On [-3,0] f'(x) >= 0 from the above graph, so f(x) is increasing in that interval.

Therefore the absolute minimum must be at x = -3

f(0) = f (-3) + integral of f'(x) from x=-3 to 0

7 = f(-3) + 4.5 (4.5 is the area between the above graph and the x axis between x = -3 and 0)

So f(-3) = 2.5

So this is the absolute minimum