finding inverse function

[spacewater]

Problem \(\displaystyle f(x) = -2y^2 +5\)

Steps
\(\displaystyle x=-2y^2+5\)

\(\displaystyle x-5 = -2y^2\)

\(\displaystyle \sqrt{x-5}=-2y\)

\(\displaystyle \frac{\sqrt{x-5}}{-2} = \frac{-2y}{-2}\)

\(\displaystyle \frac{\sqrt{x-5}}{-2} = y\)

\(\displaystyle f^{-1}(x)= - \frac{\sqrt{x-5}}{2}\)


Can someone point out what I did wrong between the steps please?

 

[Loren]

Is there an error in your original statement? The expression "-2y[sup:1iprwrqc]2[/sup:1iprwrqc]+5 is a function of y, not a function of x. A function of x would have an x someplace on the right side. Maybe I'm missing something???

 

[Deleted member 4993]

Problem I am assuming actual form is:\(\displaystyle f(x) = -2x^2 +5\)

Steps (switching x & y)

\(\displaystyle x=-2y^2+5\)

\(\displaystyle x-5 = -2y^2\)
___________________________________
\(\displaystyle 5 - x = 2y^2\)

\(\displaystyle \frac{5-x}{2} \, = \, y^2\)

\(\displaystyle y \, = \, \sqrt{\frac{5-x}{2}}\)

________________________________________
\(\displaystyle \sqrt{x-5}=-2y\)

\(\displaystyle \frac{\sqrt{x-5}}{-2} = \frac{-2y}{-2}\)

\(\displaystyle \frac{\sqrt{x-5}}{-2} = y\)

\(\displaystyle f^{-1}(x)= - \frac{\sqrt{x-5}}{2}\)


Can someone point out what I did wrong between the steps please?

 

[spacewater]

Problem I am assuming actual form is:\(\displaystyle f(x) = -2x^2 +5\)

Steps (switching x & y)

\(\displaystyle x=-2y^2+5\)

\(\displaystyle x-5 = -2y^2\)
___________________________________
\(\displaystyle 5 - x = 2y^2\)

\(\displaystyle \frac{5-x}{2} \, = \, y^2\)

\(\displaystyle y \, = \, \sqrt{\frac{5-x}{2}}\)

________________________________________
\(\displaystyle \sqrt{x-5}=-2y\)

\(\displaystyle \frac{\sqrt{x-5}}{-2} = \frac{-2y}{-2}\)

\(\displaystyle \frac{\sqrt{x-5}}{-2} = y\)

\(\displaystyle f^{-1}(x)= - \frac{\sqrt{x-5}}{2}\)


Can someone point out what I did wrong between the steps please?

The correct answer is \(\displaystyle f^{-1}(x) =\frac{\sqrt{-2(x-5)}}{2}\) according to my answer sheet.... is there typo on my answer sheet?

 

[Loren]

\(\displaystyle y=-2x^2+5\)
\(\displaystyle x=-2y^2+5\)
\(\displaystyle -2y^2=x-5\)
\(\displaystyle y^2=\frac{x-5}{-2}\)
\(\displaystyle y^2=\frac{5-x}{2}\)
\(\displaystyle y^2=\frac{2(5-x)}{4}\)
\(\displaystyle y=\pm\frac{\sqrt{2(5-x)}}{2}\)
That's the way I would do it. Now, we need to figure out why we discard the negative sign.
Please note that this answer is a different form of your answer book answer, except for the plus/minus sign.