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.
