g(x)=3x-(12/X^2)-2?

Question

State the intervals where the function is increasing or decreasing and find relative extrema points: Please Help I am stuck

Answer 1

To find increasing and decreasing intervals, use the first derivative:

g'(x)=3+(24/x^3)

Now find where this is equal to 0 or where the derivative does not exist.

0=3+(24/x^3)
-3=24/x^3
-3x^3=24
x^3=-8
x^3=-2

Also the derivative does not exist when x=0

Make a chart: Test values in each interval in the first derivative.

Interval f'(x) Inc/Dec
(-infty, -2) + Inc
(-2, 0) - Dec
(0, infty) + Inc

Now we can determine the extrema from the chart. Inc to Dec means a max, so a max at x=-2. Dec to Inc means a min, so a min at 0. Now plug those values into the original function to find the y values.

Answer 2

g(x) = 3x - (12/x^2) - 2
or
g(x) = 3x - 12x^(-2) - 2

g' = 3 + 24x^(-1)

set that = 0 and solve:

3 + 24x^(-1) = 0
24/x = -3
24 = -3x
-8 = x

check a few numbers on either side of that:

g'(-6) = 3 + 24/(-6) = -1 = negative, so the graph is decreasing on the right side of x = -8
g'(-12) = 3 + 24/(-12) = 1 = positive, so the graph is increasing on the left side of x = -8