# Thread: Finding f^-1(x)

1. ## Finding f^-1(x)

Let . Find .

make its y=15x^3-3
y+3=15x^3

I think up to this point I am right,

My final answer is this y= but its wrong

2. ## Re: Finding f^-1(x)

Originally Posted by M670
Let . Find .

make its y=15x^3-3
y+3=15x^3

I think up to this point I am right,

My final answer is this y= but its wrong
When finding an inverse function, the x and y values swap. So if \displaystyle \displaystyle \begin{align*} f(x) \end{align*} is given by \displaystyle \displaystyle \begin{align*} y = 15x^3 - 3 \end{align*} then \displaystyle \displaystyle \begin{align*} f^{-1}(x) \end{align*} will be defined by

\displaystyle \displaystyle \begin{align*} x &= 15y^3 - 3 \\ x+3 &= 15y^3 \\ \frac{x + 3}{15} &= y^3 \\ \sqrt[3]{\frac{x + 3}{15}} &= y \end{align*}

So the inverse function is \displaystyle \displaystyle \begin{align*} f^{-1}(x) = \sqrt[3]{\frac{x + 3}{15}} \end{align*}

3. ## Re: Finding f^-1(x)

I just tired that answer and the program is still telling me I am wrong??

4. ## Re: Finding f^-1(x)

You must be entering it wrong then. Try ((x + 3)/(15))^(1/3).

5. ## Re: Finding f^-1(x)

thank you I entered it like 3sqrt((x+3)/15) which I guess was telling it 3 times the sqrt

6. ## Re: Finding f^-1(x)

Originally Posted by M670
thank you I entered it like 3sqrt((x+3)/15) which I guess was telling it 3 times the sqrt
cube root, not square root.