Inverse of the function: Find the range of: f(x) = 2x / sqrt(x^2 - 9)

[frank789]

Hi all

Having trouble with a question in my homework set.

Find the range of:
f(x) = 2x / sqrt(x^2 - 9)

Did I do this right?

Find inverse so,

y = 2x / sqrt(x^2 - 9)

y(sqrt(x^2 - 9)) = 2x

sqare both sides

expand left side, factor x2 out of left, divide both sides by non y factor

gives

x^2 = 9y^2 / (y^2 - 2) ---> then sqrt root both sides for x by itself and switch x and y. Need a non zero, positive denominator so x E (-inf, -2) U (2, inf) for domain of inverse which is inherently the range of the function.

Should I be worried that I squared early in the equation? Im not sure how do deal with the implications of that.

And insight appreciated!

 

[tkhunny]

You should be worried about any step that is not 100% reversible.

An Inverse will be a reflection across the line x = y. What piece does that?

Notice that the denominator, \(\displaystyle \sqrt{stuff} > 0\). Thus, x and y MUST have the same sign. If you get anything in Quadrants II or IV, something went wrong.

 

[frank789]

You should be worried about any step that is not 100% reversible.

An Inverse will be a reflection across the line x = y. What piece does that?

Notice that the denominator, \(\displaystyle \sqrt{stuff} > 0\). Thus, x and y MUST have the same sign. If you get anything in Quadrants II or IV, something went wrong.

yeah thats what I figured.

like you said graphically of the reflection over y = x, it must stay in quadrants 1 and 3 I understand that

so taking that further, negative input (because of single variable in numerator and forced positive in denominator) negative input has to map to negative output and positive input has to map to positive output. (stay in the correct quadrants as determined graphically)

I feel like im taking good steps, but in a dark room not knowing where its going. in this particular case is the squaring irrelevant? if so could you give a case where it is relevant and will lead to an incorrect answer?

 

[tkhunny]

Squaring is always relevant.

\(\displaystyle y = \dfrac{2x}{\sqrt{x^2 - 9}}\)

y can by greater than or less than zero.

\(\displaystyle y^{2} = \dfrac{4x^{2}}{x^2 - 9}\)

y^2 is positive.

All I did was square it. It makes quite a difference.

You do seem to have the right idea.

 

[mmm4444bot]

\(\displaystyle y = \dfrac{2x}{\sqrt{x^2 - 9}}\)

y can by greater than or less than or zero.

y cannot be zero :cool:

 

[tkhunny]

y cannot be zero :cool:

Extra "or".

 

[lookagain]

\(\displaystyle y = \dfrac{2x}{\sqrt{x^2 - 9}}\)

y cannot be zero :cool:

Regarding the range of the original function, y cannot belong to the interval [-2, 2].