MHB Range of Rational Functions....1

[mathdad]

Finding the range of functions can be complicated. Rational functions can be complicated. I have a terribly hard time finding the range of functions.

Find the range of y = 1/(x + 4).

How is this done?

I want the steps.

 

[MarkFL]

Shifting a function left/right won't affect its range...the given function will then have the same range as:

$$y=\frac{1}{x}$$

If you are familiar with the graph of this function, its range is readily apparent. If not, sometimes we can find the inverse function, whose domain is the range of the original. The function above is special in that it is its own inverse. So, take the domain of the above function, and you have its range as well, which will be the range of the original function you gave. :D

 

[mathdad]

y = 1/(x + 4)

Set x + 4 = 0.

x + 4 - 4 = -4

x = -4

The domain is ALL REALL NUMBERS except that x cannot be -4.

If I understood correctly, the range is ALL REAL NUMBERS except for -4.

Is this right?

 

[MarkFL]

y = 1/(x + 4)

Set x + 4 = 0.

x + 4 - 4 = -4

x = -4

The domain is ALL REALL NUMBERS except that x cannot be -4.

If I understood correctly, the range is ALL REAL NUMBERS except for -4.

Is this right?

I was talking about the following function:

$$y=\frac{1}{x}$$

being its own inverse. :D

 

[mathdad]

For y = 1/x, the domain is ALL REALL NUMBERS except for 0.

So, the range of y = 1/( x + 4) is ALL REAL NUMBERS except for 0.

Yes? No?

 

[mathdad]

Are you saying that I need to find the inverse of
y = 1/(x + 4)?

Let me see.

x = 1/(y + 4)

Then y + 4 = 0.

y + 4 - 4 = -4

y = -4

The domain of the inverse function is ALL REAL NUMBERS except for -4.

This is the range of the given function.

Correct?

 

[MarkFL]

For y = 1/x, the domain is ALL REALL NUMBERS except for 0.

So, the range of y = 1/( x + 4) is ALL REAL NUMBERS except for 0.

Yes? No?

Yes, that's right.

Are you saying that I need to find the inverse of
y = 1/(x + 4)?

I'm saying finding the inverse and taking its domain as the range of the original is one way to go about it. :D

Let me see.

x = 1/(y + 4)

Now you want to solve for $y$, and then you have the inverse function.

 

[mathdad]

x = 1/(y + 4)

x(y + 4) = 1

xy + 4y = 1

y(x + 4) = 1

y = 1/(x + 4)

Like you said, it is its own inverse.

The domain of y = 1/(x + 4) is ALL REAL NUMBERS except for -4.

 

[MarkFL]

x = 1/(y + 4)

x(y + 4) = 1

xy + 4y = 1

y(x + 4) = 1

y = 1/(x + 4)

Like you said, it is its own inverse.

The domain of y = 1/(x + 4) is ALL REAL NUMBERS except for -4.

Let's go back to:

$$x=\frac{1}{y+4}$$

This means:

$$y+4=\frac{1}{x}$$

Or:

$$y=\frac{1}{x}-4$$

This demonstrates that the shift left/right has no bearing on the range.

Can you spot your algebraic error when solving for $y$ to get the inverse?

 

[mathdad]

Let me start again.

y = 1/(x + 4)

x = 1/(y + 4)

x(y + 4) = 1

xy + 4x = 1

xy = 1 - 4x

y = (1 - 4x)/x

The domain of the inverse is ALL REAL NUMBERS except for 0.

This is the range of the original function.