Finding range for a function

C

Chris*

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Jan 9, 2007
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Hello, one of the problems in my homework asks to find the domain and range of this function:
\(\displaystyle f(x)=\frac{1}{\sqrt[]{x^{2}-4}}\)

For the domain, I set x[sup:mleob1nm]2[/sup:mleob1nm]-4 >0 and got Domain\(\displaystyle \left(-\infty,-2 \right)\cup\left(2,\infty \right)\)

How do I algebraically find the range?
 


Hi Chris:

With this type of function, there is no algebraic "formula" for determining the range.

I'm thinking of two ways to determine the range (without looking at a graph); one approach is numerical, and the other is analytical.

The analytical approach is easier, because you can reason it out in your head in less than a minute, but only if you're familiar with things like: the behavior (i.e., range) of the square root function; the graph of the reciprocal function y=1/x or other rational functions; asymptotes; infinity.

(I'll discuss a numerical approach at the end.)

We know that the denominator sqrt(x^2 -4) is always positive, for the domain you determined, because the square root of a positive is always positive.

Also, as the absolute value of x increases, so does the value of sqrt(x^2 - 4). [Think of the behavior of sqrt(x) as x increases.]

In other words, as the value of x gets farther away from -2 moving in the negative direction, or farther away from +2 moving in the positive direction, the denominator of the given ratio is growing without bound.

But the numerator is fixed at 1.

In any ratio where the numerator does not change and the denominator continues to grow, the value of the overall ratio approaches zero. In fact, as x become infinitely large (in absolute value), the value of f(x) is continually getting closer and closer to zero, without ever reaching zero.

In other words, we can envision the graph of f(x) getting closer and closer to the x-axis, without ever reaching it.

And since we know that the denominator is always positive, the value of the ratio will never become negative.

In other words, we know that the graph of f(x) will never "jump" the x-axis into negative "territory".

That takes care of the bottom end of the range. f(x) must be > 0.

Now think of x moving in the positive direction (away from negative infinity) toward -2 OR moving in the negative direction (away from positive infinity) toward +2. We have the opposite situation.

The denominator sqrt(x^2 - 4) is continually getting closer to zero. (Again, think of the square root function, as x gets closer to zero.)

In any ratio where the numerator does not change and the denominator continues to shrink, the value of the overall ratio becomes infinitely large, in absolute value.

This takes care of the upper end of the range. f(x) is growing without bound.

Therefore, the range is (0 , infinity).

Okay, the second approach is numerical.

Make a chart to see the behavior of f(x).

Pick values of x that are interesting and relevant.

Use a machine to help you, if you have one.

You'll investigate entries like (scroll):

Code: 
         x             f(x)

+/- 2.000000001     15811.38830

+/- 2.00001         158.1138830

+/- 2.01            4.993761696

+/- 2.1             1.561737619

+/- 3               0.447213596

+/- 5               0.218217890

+/- 10              0.102062073

+/- 100             0.010002001

+/- 10000           0.000100000

See the trends? Hmmm. Perhaps, it's good to know about the behavior of 1/x before taking the numerical approach, too. Then, the range would become obvious, with such a chart.

Of course, if you have access to a graph of f(x), then you have a picture of the range. That's the easiest!

Cheers,

~ Mark 8-)

 
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