finding inverse function

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spacewater

Junior Member
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Jul 10, 2009
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Problem f(x)=−2y2+5\displaystyle f(x) = -2y^2 +5

Steps
x=−2y2+5\displaystyle x=-2y^2+5

x−5=−2y2\displaystyle x-5 = -2y^2

x−5=−2y\displaystyle \sqrt{x-5}=-2y

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

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

f−1(x)=−x−52\displaystyle f^{-1}(x)= - \frac{\sqrt{x-5}}{2}


Can someone point out what I did wrong between the steps please?
 
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???
 
spacewater said:
Problem I am assuming actual form is:f(x)=−2x2+5\displaystyle f(x) = -2x^2 +5

Steps (switching x & y)

x=−2y2+5\displaystyle x=-2y^2+5

x−5=−2y2\displaystyle x-5 = -2y^2
___________________________________
5−x=2y2\displaystyle 5 - x = 2y^2

5−x2 = y2\displaystyle \frac{5-x}{2} \, = \, y^2

y = 5−x2\displaystyle y \, = \, \sqrt{\frac{5-x}{2}}

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

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

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

f−1(x)=−x−52\displaystyle f^{-1}(x)= - \frac{\sqrt{x-5}}{2}


Can someone point out what I did wrong between the steps please?
Click to expand...
 
Subhotosh Khan said:
spacewater said:
Problem I am assuming actual form is:f(x)=−2x2+5\displaystyle f(x) = -2x^2 +5

Steps (switching x & y)

x=−2y2+5\displaystyle x=-2y^2+5

x−5=−2y2\displaystyle x-5 = -2y^2
___________________________________
5−x=2y2\displaystyle 5 - x = 2y^2

5−x2 = y2\displaystyle \frac{5-x}{2} \, = \, y^2

y = 5−x2\displaystyle y \, = \, \sqrt{\frac{5-x}{2}}

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

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

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

f−1(x)=−x−52\displaystyle f^{-1}(x)= - \frac{\sqrt{x-5}}{2}


Can someone point out what I did wrong between the steps please?
Click to expand...
Click to expand...


The correct answer is f−1(x)=−2(x−5)2\displaystyle f^{-1}(x) =\frac{\sqrt{-2(x-5)}}{2} according to my answer sheet.... is there typo on my answer sheet?
 
y=−2x2+5\displaystyle y=-2x^2+5
x=−2y2+5\displaystyle x=-2y^2+5
−2y2=x−5\displaystyle -2y^2=x-5
y2=x−5−2\displaystyle y^2=\frac{x-5}{-2}
y2=5−x2\displaystyle y^2=\frac{5-x}{2}
y2=2(5−x)4\displaystyle y^2=\frac{2(5-x)}{4}
y=±2(5−x)2\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.
 
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