Question

(x+3)(x−2)/(x+7)^2 ≥0

Answer 1

(x+3)(x−2)/(x+7)^2 ≥0 when x ≤ -3 or x ≥ 2.

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.