How do you do this question?

A) x ≤ -2 and 0 ≤ x ≤ 3

Step-by-step explanation:

g(x) is decreasing when g'(x) is negative.

Use second fundamental theorem of calculus to find g'(x).

g(x) = ∫₋₁ˣ (t³ − t² − 6t) / √(t² + 7) dt

g'(x) = (x³ − x² − 6x) / √(x² + 7) (1)

To find when g'(x) is negative, first find where it is 0.

0 = (x³ − x² − 6x) / √(x² + 7)

0 = x³ − x² − 6x

0 = x (x² − x − 6)

0 = x (x − 3) (x + 2)

x = -2, 0, or 3

Check the intervals before and after each zero.

x < -2, g'(x) < 0

-2 < x < 0, g'(x) > 0

0 < x < 3, g'(x) < 0

3 < x, g'(x) > 0

g(x) is decreasing on the intervals x ≤ -2 and 0 ≤ x ≤ 3.

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