Answer:
(-∞, -1-(2/3)√2] ∪ [-1+(2/3)√2, ∞)
Step-by-step explanation:
To make it easier to differentiate, we'll rewrite the function as ...
h(x) = 2/(x-5) -1/(x-2)
Then the derivative is ...
h'(x) = -2/(x-5)^2 +1/(x-2)^2
This will be zero when ...
-2(x-2)^2 +(x-5)^2 = 0
-2(x^2 -4x +4) +(x^2 -10x +25) = 0
-x^2 -2x +17 = 0
x^2 +2x +1 = 17 +1
x +1 = ±√18 = ±3√2
x = -1 ±3√2
The values of the function at these locations are ...
h(-1-3√2) = -1 +(2/3)√2 ≈ -1.9428
h(-1+3√2) = -1 -(2/3)√2 ≈ -0.0572
Then the range of h(x) is ...
(-∞, -1-(2/3)√2] ∪ [-1+(2/3)√2, ∞)
