Find the Inverse function (Math Help)

[TamaraR]

how do I find the range of the function for the equation.... y=3x^3*1/x^2

 

[ksdhart]

I'm sorry, but I'm a bit confused right now. In your title, you say you need help finding the inverse of a function. But then in your body, you say you need to find the range of the function. So, which is it? We can't help you unless we know what the problem is.

 

[Math Doggie]

how do I find the range of the function for the equation.... y=3x^3*1/x^2

I'm assuming you wanted to get the inverse function and determine the domain of the inverse function.

Your equation simplifies to a linear function so the inverse function will be a linear function too. The domain of a linear function is always all real numbers so D = -∞ to ∞.

Your equation simplifies like this ->

y=3x^3*1/x^2 =>

y=3x^3/x^2 =>

y=3x

The inverse function is calculated like this ->

y=3x

x=3y

y=x/3

You can prove that y=3x and y=x/3 are inverse functions by plugging either equation into the x value of the other function. It should solve to x.

 

[Ishuda]

I'm assuming you wanted to get the inverse function and determine the domain of the inverse function.

Your equation simplifies to a linear function so the inverse function will be a linear function too. The domain of a linear function is always all real numbers so D = -∞ to ∞. ...

Strictly speaking, because of the way the function is written it has a (removable) singularity and the domain D of the original function is D={(-∞, ∞) - {0}} and similarly for the range. Thus there are similar restrictions on the inverse function.

At least that's the way I would answer the question.

 

[Math Doggie]

Strictly speaking, because of the way the function is written it has a (removable) singularity and the domain D of the original function is D={(-∞, ∞) - {0}} and similarly for the range. Thus there are similar restrictions on the inverse function.

At least that's the way I would answer the question.

Excellent point, I left that out. Thanks for the help.

Since x^2 is in the denominator of the equation, a value of x = 0 would cause the equation to be undefined, so zero can't be included in the domain. Technically, there would be hole at x = 0 in the equation. The range would be similarly affected.

So, in interval notation -> D = (-∞, 0) U (0, ∞) and R = (-∞, 0) U (0, ∞).

Have I got that right?

 

[Deleted member 4993]

I suspect the OP meant to write:

y = 3x³ + 1/x²

 

[Ishuda]

Excellent point, I left that out. Thanks for the help.

Since x^2 is in the denominator of the equation, a value of x = 0 would cause the equation to be undefined, so zero can't be included in the domain. Technically, there would be hole at x = 0 in the equation. The range would be similarly affected.

So, in interval notation -> D = (-∞, 0) U (0, ∞) and R = (-∞, 0) U (0, ∞).

Have I got that right?

Yes, I would agree it can be written that way and, in fact, may be a better notation that the one I used for some people.