Question #a678e

Sep 29, 2017

Explanation:

...understand that an inverse function (call it g(x)) meets the condition:

$f \left(g \left(x\right)\right) = x$

...so we can just plug g(x) into the formula we are given for f(x), and then set that equal to x, and then algebraically solve for it. It will look like:

$\frac{2 \left(g \left(x\right)\right) + 1}{4 \left(g \left(x\right)\right) - 1} = x$

...multiply both sides by$\left(4 \left(g \left(x\right)\right) - 1\right)$, giving:

$2 \left(g \left(x\right)\right) + 1 = x \left(4 \left(g \left(x\right)\right) - 1\right) = 4 x \left(g \left(x\right)\right) - x$

...we want to get the terms involving g(x) all on the same side,
so we subtract 2(g(x)) from both sides:

$1 = 4 x \left(g \left(x\right)\right) - x - 2 \left(g \left(x\right)\right)$

we want terms NOT involving g(x) all on the same side, so we add $x$ to both sides:

$1 + x = 4 x \left(g \left(x\right)\right) - 2 \left(g \left(x\right)\right)$

...now we can factor out g(x) on the right side:

$1 + x = g \left(x\right) \left(4 x - 2\right)$

...and now we can divide both sides by (4x-2), and we've got it!

$\frac{1 + x}{4 x - 2} = g \left(x\right)$

...check your work when you can. Yeah, I know, it's double the work, but it will save you on occasion, and will show your instructor that you care. He'll write you a nice recommendation or something.

We check this result by plugging in our derived function g(x) into the original function definition, and show that f(g(x)) = x.

$\frac{2 \left(\frac{1 + x}{4 x - 2}\right) + 1}{4 \left(\frac{1 + x}{4 x - 2}\right) - 1}$

$= \frac{\frac{2 + 2 x}{4 x - 2} + \frac{4 x - 2}{4 x - 2}}{\frac{4 + 4 x}{4 x - 2} - \frac{4 x - 2}{4 x - 2}}$

$= \frac{\frac{6 x}{4 x - 2}}{\frac{6}{4 x - 2}} = \left(\frac{6 x}{4 x - 2}\right) \cdot \left(\frac{4 x - 2}{6}\right)$

$= \frac{6 x}{6} = x$

...they GOTTA be happy with that!

GOOD LUCK!