Answer:
(x+3)(x−2)/(x+7)^2 ≥0 when x ≤ -3 or x ≥ 2.
Step-by-step explanation:
(x+3)(x−2)/(x+7)^2 ≥0
The critical values are (x + 3)(x - 2) = 0.
That is when x = -3 and x = 2.
Set up a table:
x < - 3 x = -3 -3 < x < 2 x = 2 x > 2
x + 3 -ve 0 +ve 0 +ve
x - 2 -Ve -ve -ve 0 +ve
(x+7)^2 +ve +ve +ve +ve +ve
Whole
function + ve 0 - ve 0 +ve
So the function is ≥ 0 when x ≤ -3 and x ≥ 2.
