Finding domain of a function.

**xxStrikeback** · September 11th 2011, 01:44 PM

Find the domain of f(x).

Then find the inverse of the function, and the domain of the inverse.

I know how to find the domain and inverse, but I just have no idea how to do it in this function.

For domain, I'm guess 3 - e^2x > or = 0

-e^2x > or = -3
e^2x < or = 3
2x < or = ln3
x < or = ln3 / 2

Is that right? Something just doesn't feel right.

EDIT: K, nvm. That first part was right.
But I didn't get the inverse right.

x^2=3-e^2y
x^2 - 3 = -e^2y
ln(x^2 - 3) = -2y
-ln(x^2 - 3) / 2 = y

**e^(i*pi)** · September 11th 2011, 01:56 PM

Originally Posted by xxStrikeback

Find the domain of f(x).

Then find the inverse of the function, and the domain of the inverse.

I know how to find the domain and inverse, but I just have no idea how to do it in this function.

For domain, I'm guess 3 - e^2x > or = 0

-e^2x > or = -3
e^2x < or = 3
2x < or = ln3
x < or = ln3 / 2

Is that right? Something just doesn't feel right.

Nothing wrong there. $x < \dfrac{\ln(3)}{2}$ is the correct domain. The range is $f(x) \geq 0$ (this will come in handy later)

For the inverse I use the method of saying that $f(x) = y$ and then solve for x in terms of y.

$y^2 = 3-e^{2x} \Leftrightarrow e^{2x} = 3-y^2$

Can you finish off?

Once done replace x with $f^{-1}(x)$ and $y$ with $x$ and find the domain in the usual way.

You can check your answer due to the fact that the domain of $f^{-1}(x)$ is the range of f(x) and vice versa.

**xxStrikeback** · September 11th 2011, 02:02 PM

Check the first post, sorry, I edited, I assume, as you were posting.

x^2=3-e^2y
x^2 - 3 = -e^2y
ln(x^2 - 3) = -2y
-ln(x^2 - 3) / 2 = y

I submitted that, but apparently it was not right.

**e^(i*pi)** · September 11th 2011, 02:10 PM

Originally Posted by xxStrikeback

Check the first post, sorry, I edited, I assume, as you were posting.

x^2=3-e^2y
x^2 - 3 = -e^2y

This bit is good

Originally Posted by xxStrikeback

ln(x^2 - 3) = -2y
-ln(x^2 - 3) / 2 = y

I submitted that, but apparently it was not right.

You need to divide by -1 before you take the logarithm since the logarithm is not defined (in the real numbers) for negative arguments and $-e^{2y}$ is always negative so if you try to do what you did you end up trying to take the log of a negative number.

$x^2-3 = -e^{2y}$

$-(x^2-3) = e^{2y} \Leftrightarrow 3-x^2 = e^{2y}$

Then continue as normal to get $f^{-1}(x) = \dfrac{1}{2}\ln(3-x^2)$

**xxStrikeback** · September 11th 2011, 02:16 PM

Cheers, got it.

Now you mention earlier that the range of f(x) was > or = 0. So this means the domain of the inverse should be [0,inf), right?

When I put this in, it tells me that the [, 0, and ) are right, but the infinity is wrong.

So I'm confused now on finding the domain of the inverse. What value can x not equal?

**e^(i*pi)** · September 11th 2011, 02:22 PM

Originally Posted by xxStrikeback

Cheers, got it.

Now you mention earlier that the range of f(x) was > or = 0. So this means the domain of the inverse should be [0,inf), right?

I think I was incorrect earlier when stating that bit, got carried away after the first sentence!

Originally Posted by xxStrikeback

When I put this in, it tells me that the [, 0, and ) are right, but the infinity is wrong.

So I'm confused now on finding the domain of the inverse. What value can x not equal?

For the domain of the inverse $3-x^2 > 0$ so $|x| < \sqrt{3}$ which is the same as $-\sqrt{3} < x < \sqrt{3}$

**xxStrikeback** · September 11th 2011, 02:35 PM

Ah makes sense. Thank you very much.

While on inverses, I have a question on using f(x) = x^3 + 4x + 2 to find f^-1(2) = ?

I'm not really sure how to get the inverse here. All I've really done is.

x=y^3 + 4y +2
x-2 = y(y^2+4)
x-2 / y = y^2 + 4

End up just coming back to the original equation, just not sure how to get it with the x, or rather, y^3.

**e^(i*pi)** · September 11th 2011, 02:55 PM

Originally Posted by xxStrikeback

Ah makes sense. Thank you very much.

While on inverses, I have a question on using f(x) = x^3 + 4x + 2 to find f^-1(2) = ?

I'm not really sure how to get the inverse here. All I've really done is.

x=y^3 + 4y +2
x-2 = y(y^2+4)
x-2 / y = y^2 + 4

End up just coming back to the original equation, just not sure how to get it with the x, or rather, y^3.

The inverse is definitely not pretty so I'm thinking there must be some kind of shortcut involved, possibly using the fact that:

$f(f^{-1}(2)) = 2$

Other than that I'm not sure I'm afraid >.<

**xxStrikeback** · September 11th 2011, 03:00 PM

Lol, here's the full problem.

Properties of Inverse Functions Theorem: Suppose f and g are inverse functions.

• The range of f is the domain of g and the domain of f is the range of g.
•f(a) = b if and only if g(b) = a.
•(a, b) is on the graph of f if and only if (b, a) is on the graph of g.

The function
f(x) = x3 + 4x + 2
is one-to-one. Since finding a formula for its inverse is beyond the scope of this textbook, use Properties of Inverse Functions Theorem to help you compute
f −1(2).

Yeah I tried to solve it by hand and then just WolframAlpha'd it and was in shock. I don't know if any of those properties can be used for this?

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