Expert: Scotto - 8/3/2009

Question
Please help me with the following problem:  Use calculus to find the intervals of increase and decrease for the function

f(x)=x^2-(1/x^2)

Answer
f(x)=x^2-(1/x^2); note that at x=0, the function is undefined.

This means that f'(x) = 2x - (2/x³).

Finding the Intervals
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Setting f'(x) to 0 gives 0 = 2x - (2/x³) implies 2x = 2/x³.

Now when we multiply both sides by x³,
it is good to note that we have to also count for when x³ = 0, which is at x = 0.

Multiplying by x³ give 2x^4 = 2.
Dividing both sides by 2 gives x^4 = 1.
From this equation, you get x = ±1.

When I note we also had look at x=0,
we get intervals to look at are (-∞,-1), (-1,0), (0,1) and (1,∞).

Notes
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Note that when x gets far from 0, in the value of f(x),
the x² terms takes over since 1/x is going to 0.
Also note that when x gets close to 0,
the –2/x² terms takes over since x is going to 0.

Intervals
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(-∞,-1)
Looking back at f'(x), if x is a large negative number, 2x is also a large number and 2/x³ is close to zero, so the result is a large negative number.  This means that the function is decreasing in (-∞,-1).

(-1,0)
If x is just barely negative, 2x is a very small negative number, but -2/x³ is negative over a really small negative, which makes a large positive.  So here, the function is increasing on (-1, 0).

(0.1)
If x is a small positive, the x^2 is a really small positive, but -2/x³ is a really large negative.  This means that in the interval (0,1), the derivative is negative and the function is decreasing on (0, 1).

(1,∞)
If x is a large positive, then 2x is also a large large positive and 1/x³ is a really really really small number, so when the second is subtracted from the first, the answer is positive, so the function is increasing on (1,∞).

Final Note:
Since both of these functions for f(x) are even, the function is symmetric about the line x = 0.