Answer:
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.
