Question #9099e

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Question #9099e

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Answer 1 · 2017-03-09T20:27:49

${f (x)}^{- 1} = \frac{x^{2}}{8} + \frac{7}{2}$ $f(x)^-1=x^2/8+7/2$

This can also be written as one fraction:

${f (x)}^{- 1} = \frac{x^{2} + 28}{8}$ $f(x)^-1 = (x^2 +28)/8$

Explanation:

Rules
When finding the inverse, $f^{- 1} (x)$ $f^(-1)(x)$ , the easiest method is to switch the variables $x$ $x$ and $y$ $y$ . So $y = f (x)$ $y=f(x)$ is now $x = f (y)$ $x=f(y)$ .
Afterwards, you need to isolate $y$ $y$ again to get the proper form of the inverse.

Application
$f (x) = y = 2 {(2 x - 7)}^{\frac{1}{2}}$ $f(x)=y=2(2x-7)^(1/2)" "$ Applying the "switch" ...

$f (y) = x = 2 {(2 y - 7)}^{\frac{1}{2}}$ $f(y)=x=2(2y-7)^(1/2)$

Solving for $y$ $y$ ...

$x = 2 \sqrt{2 y - 7}$ $x=2\sqrt{2y-7}$

$\frac{1}{2} x = \sqrt{2 y - 7}$ $1/2x=\sqrt{2y-7}$

${(\frac{1}{2} x)}^{2} = 2 y - 7$ $(1/2x)^2=2y-7$

$\frac{1}{4} x^{2} + 7 = 2 y$ $1/4x^2+7=2y$

$\frac{1}{2} (\frac{1}{4} x^{2} + 7) = y$ $1/2(1/4x^2+7)=y$

$\frac{1}{8} x^{2} + \frac{7}{2} = y$ $1/8x^2+7/2=y$

Confused? Feel free to drop a comment and tell me where you're still not understanding things! :)

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Question #9099e

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