How do you determine the intervals where each of the functions is increasing, and decreasing of the function:?

Question

g(x) = x + 1/x^2

Answer 1

g(x) = x + 1/x^2

It's best to convert this to a more desirable form; in this case, it would be a single fraction.

g(x) = x^3/x^2 + 1/x^2

Now, merge into one fraction.

g(x) = [x^3 + 1]/x^2

To find the intervals of increase/decrease, we must take the derivative and then obtain our critical values. Using the quotient rule,

g'(x) = ( [3x^2][x^2] - [x^3 + 1][2x] ) / x^4

Simplify the top,

g'(x) = ( 3x^4 - 2x^4 - 2x ) / (x^4)
g'(x) = (x^4 - 2x)/x^4

Factor out an x,

g'(x) = x(x^3 - 2)/x^4

Cancel an x out.

g'(x) = (x^3 - 2)/x^3.

Make g'(x) = 0.

0 = (x^3 - 2)/x^3

Critical values are defined to be where f'(x) = 0 or where f'(x) is undefined. The solution to the above equation that makes it 0 is when the numerator is 0; the above equation is undefined when the denominator is 0.

Equating the numerator to 0, we get x^3 - 2 = 0, which means
x = 2^(1/3) {the cube root of 2}.

Equating the denominator to 0, we have x^3 = 0, which means x = 0.

So our critical values are {0, 2^(1/3)}

At this point, draw a number line consisting of our critical values.

. . . . . . . . . (0). . . . . . . . . . . . .(2^(1/3)) . . . . . . . .

For f'(x), test a single value in each region around the critical values. If that single value is positive, mark the region as positive; if it is negative, mark the region as negative.

Test f'(-1): then (x^3 - 2)/x^3 = (-1 - 2)/(-1)^3 = (-3)/(-1) = 3, which is positive. Mark the region as positive.

. . . . {+}. . . . (0). . . . . . . . . . . . .(2^(1/3)) . . . . . . . .

Test f'(1) {since 1 lies between 0 and 2^(1/3)}. Then
f'(1) = (1^3 - 2)/(1^3) = (1 - 2)/(1) = -1, which is negative. Mark the region as negative.

. . . . {+}. . . . (0). . . . . .{-}. . . . . .(2^(1/3)) . . . . . . . .

Remember that we can test ANY single value within the region. For the 3rd and final region, I'm going extreme and testing a very large number, 10000.

For f'(10000), (x^3 - 2)/x^3 = [really large positive number] divided by [really large positive number], which gives the result of positive. Mark the region as positive.

. . . . {+}. . . . (0). . . . . .{-}. . . . . .(2^(1/3)) . . . {+} . . . .

The regions that are positive are intervals where f(x) is increasing; the regions that are negative are intervals where f(x) is decreasing. Be sure to include the critical number that makes f'(x) equal to 0, and exclude the critical number that makes f'(x) undefined.

Intervals of increase: (-infinity, 0) U [2^(1/3), infinity)
Intervals of decrease: (0, 2^(1/3)]

{Note the square brackets around the critical value that made f'(x) = 0 and the round brackets around the critical value making it undefined.}

*****
Going one step further, you can deduce that a local minimum occurs at x = 2^(1/3) (since the function is decreasing until that point, and then increasing). This means we can solve for the local minimum.

f(2^(1/3)) = [ [2^(1/3)]^3 + 1 ] / [ 2^(1/3)]^2

= [ 2 + 1 ] / [2^(2/3)]
= 3/[2^(2/3)]

Local minimum at (2^(1/3) , 3/2^(2/3) )

Answer 2

First you must find the derivative of the function.
After that you must find out where the derivative equals zero, which will give you your possible relative extrema.
Then you determine whether the derivative is positive or negative around the possible extrema. If the derivative is negative the function is decreasing and if the derivative is positive then the function is increasing.