Finding critical points and absolute extrema

Question

Question Image Link: https://i.imgur.com/vFvf65s.png

The graph of g(t) is shown on the interval [0, 6] and f(x) = ∫ 0 to x g(t) dt.

a) Find the critical points of f(x), and identify each as a relative min, max, or neither.

b) Find the absolute extrema of f(x).

Answer 1

By the Fundamental Theorem of calculus,

g(x) = f'(x)

(a)

The critical points of f(x) are where its derivative, g(x), is 0.

So they are x = 2 and x = 4.

They are relative maxima or minima depending on the polarity of f"(x),

which is g'(x).

g'(2) = 1, therefore x = 2 is local minimum.

g'(4) = -1, therefore x = 4 is local maximum.

(b)

Absolute extrema are either at critical points or boundary of domain.

- Absolute maximum could be at x = 4, x = 0, or x = 6.

It cannot be x = 6 because f(4) > f(6).

The reason is that for x > 4 g(x) < 0, and therefore f(x) is decreasing there.

Absolute maximum is actually at x = 0, because f(0) > f(4) for the following reason.

f(4) = f(0) + ∫₀² g(t)dt + ∫₂⁴g(t)dt

∫₂⁴g(t)dt = 1 because it is the area of the triangle with base 2 and height 1.

∫₀² g(t)dt < -1 because the rectangle between 0 and 1 alone has area -1.

- Absolute minimum could be at x = 2 or x = 6.

It is at x = 6 for the analogous reason as above.

f(6) = f(2) + ∫₂⁴g(t)dt + ∫₄⁶g(t)dt

∫₂⁴g(t)dt = 1, and ∫₄⁶g(t)dt < -1 because the rectangle between 5 and 6 alone has area -1.