Question

How do you do this question?

Answer 1

Answer:

0.471

Step-by-step explanation:

g'(x) = 0 at the relative maximum.  To find g'(x), use the second fundamental theorem of calculus.

g(x) = ∫₋₁ˣ (-½ + cos(t³ + 2t)) dt

g'(x) = -½ + cos(x³ + 2x)

Setting equal to 0 and solving:

0 = -½ + cos(x³ + 2x)

cos(x³ + 2x) = ½

x³ + 2x = π/3 or 5π/3

x = 0.471 or 1.360

At a relative maximum, g'(x) changes signs from positive to negative.  Test points on both sides of each critical value.

g'(0) = 0.5

g'(1) = -1.490

g'(π/2) = 0.242

At x = 0.471, g'(x) changes signs from positive to negative.

At x = π/2, g'(x) changes signs from negative to positive.

So x = 0.471 is a relative maximum, and x = 1.360 is a relative minimum.