Finding f^-1(x)

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October 4th 2012, 05:48 PM
M670

Finding f^-1(x)

Let http://webwork.mathstat.concordia.ca...19f0de6441.png. Find http://webwork.mathstat.concordia.ca...59fc9bac51.png. http://webwork.mathstat.concordia.ca...59fc9bac51.png

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=http://webwork.mathstat.concordia.ca...906e017fc1.png but its wrong
October 4th 2012, 05:51 PM
Prove It

Re: Finding f^-1(x)

Quote:

Originally Posted by M670

Let http://webwork.mathstat.concordia.ca...19f0de6441.png. Find http://webwork.mathstat.concordia.ca...59fc9bac51.png. http://webwork.mathstat.concordia.ca...59fc9bac51.png

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=http://webwork.mathstat.concordia.ca...906e017fc1.png but its wrong

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

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

So the inverse function is $\displaystyle \begin{align*} f^{-1}(x) = \sqrt[3]{\frac{x + 3}{15}} \end{align*}$
October 4th 2012, 06:02 PM
M670

Re: Finding f^-1(x)

I just tired that answer and the program is still telling me I am wrong??
October 4th 2012, 06:07 PM
Prove It

Re: Finding f^-1(x)

You must be entering it wrong then. Try ((x + 3)/(15))^(1/3).
October 4th 2012, 06:10 PM
M670

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
October 5th 2012, 03:20 AM
skeeter

Re: Finding f^-1(x)

Quote:

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.