I really need help with my homework. Thank you!

Question

The graph of y = f ′(x), the derivative of f(x), is shown below. Given f(4) = 6, evaluate f(0).

Options:

1)	-2
2)	2
3)	4
4)	10

Answer 1

Hi Kehlani F.,

I relied on your description of the behavior of f ' (x), not being able to open the figure . Since you stated that there were exactly 3 segments, then the f ' (x) graph looks like a "linearized sine" wave, which must be symmetric to rotation by 180 ^o around (0,0). But the description does not state what are the minima and maxima of that graph; therefore I must assume that they are unspecified. The graph of f (x) will therefore consist of 3 parabolic segments; it is concave downwards from x = -4 to -2 and from 2 to 4, and concave upwards from x= -2 to 2 . The graph has zero slope at -4, 0, and 4. However, the actual amplitude (minimum to (maximum=6) ) in this interval is unknown. So all one can say is that f(0) is less than f(4), i.e. it might be either 2 or -2 .

If the description of "the graph" is of f(x) rather than of f '(x), then the above analysis does not apply. But why ask the question about f(0) if you have the graph of f(x)?

Nitin P. appears to me to have misinterpreted the given data on the behavior of f '(x). If it is in 2 segments over the x=0 to 4 interval (the left segment being only half of the segment from x = -2 to 2 ), then it must reverse slope at x=2 in order to have value 0 at both x = 0 and 4 . That makes for 2 triangles that must be integrated, not one. Quite a different condition!

Now, if you DO have the values of f '(x) at -2 and 2, then you can solve exactly, of course.

-- Cheers, -- Mr. d.

Answer 2

Since you have the graph of f'(x), you can use the Fundamental Theorem of calculus:

∫₀⁴ f'(x) dx = f(4) - f(0)

f(0) = f(4) - ∫₀⁴ f'(x) dx

The integral is just the area under the curve. Since the curve in this interval is just a triangle, the area is

A = ∫₀⁴ f'(x) dx = (1/2)(4)(2) = 4

Therefore, the answer is

f(0) = 6 - 4 = 2