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.
The right answer is b
Step-by-step explanation:
You do you do the calculations you get that answer
your answer for this question would be A)
You would need to draw a graph or get some graph paper and plot those points. Then calculate the rise and the run. Rise is over the run. Then that is your slope. You get the rise by counting how many units up or down it takes to get to the point, and for the run, you count how many units across it is to get to the point. don't forget, if your units are going in a negative direction, then the number will be negative.
The answer selected in the question is correct E. is the answer
