# Given f(x) = (3-2x) / (2x+1) and f(g(x)) = 7 - 3x how do you find g(x)?

##### 1 Answer
Feb 4, 2016

$g \left(x\right) = \frac{4 - 3 x}{- 16 + 6 x}$

#### Explanation:

$f \left(g \left(x\right)\right)$ can be computed by plugging $g \left(x\right)$ for every occurence of $x$ in $f \left(x\right)$.

Even though we don't know $g \left(x\right)$ yet, we can still do this:

$f \left(x\right) = \frac{3 - 2 x}{2 x + 1} \text{ "=> " } f \left(g \left(x\right)\right) = \frac{3 - 2 g \left(x\right)}{2 g \left(x\right) + 1}$

We also know that $f \left(g \left(x\right)\right) = 7 - 3 x$, so we have:

$\frac{3 - 2 g \left(x\right)}{2 g \left(x\right) + 1} = 7 - 3 x$

Let me write $g$ instead of $g \left(x\right)$ for better readability:

$\frac{3 - 2 g}{2 g + 1} = 7 - 3 x$

Now, you need to solve this equation for $g$:

... multiply both sides with $\left(2 g + 1\right)$...

$\iff 3 - 2 g = \left(7 - 3 x\right) \cdot \left(2 g + 1\right)$

$\iff 3 - 2 g = \left(7 - 3 x\right) \cdot 2 g + \left(7 - 3 x\right)$

Bring all products that include $g$ to the left side and everything else to the right side.
So, subtract $\left(7 - 3 x\right) \cdot 2 g$ on both sides, and subtract $3$ on both sides:

$\iff - 2 g - \left(7 - 3 x\right) \cdot 2 g = \left(7 - 3 x\right) - 3$

... factorize $g$ on the left side...

$\iff \left(- 2 - 14 + 6 x\right) \cdot g = 4 - 3 x$

$\iff \left(- 16 + 6 x\right) \cdot g = 4 - 3 x$

... divide both sides by $\left(- 16 + 6 x\right)$...

$\iff g = \frac{4 - 3 x}{- 16 + 6 x} = \frac{4 - 3 x}{2 \left(- 8 + 3 x\right)}$

Thus, we have

$g \left(x\right) = \frac{4 - 3 x}{- 16 + 6 x}$

It might be a good idea to test if the calculation was correct. To do so, compute $f \left(g \left(x\right)\right)$:

$f \left(g \left(x\right)\right) = f \left(\frac{4 - 3 x}{- 16 + 6 x}\right)$

$= \frac{3 - 2 \cdot \frac{4 - 3 x}{2 \left(- 8 + 3 x\right)}}{2 \cdot \frac{4 - 3 x}{2 \left(- 8 + 3 x\right)} + 1}$

$= \frac{3 - \frac{4 - 3 x}{- 8 + 3 x}}{\frac{4 - 3 x}{- 8 + 3 x} + 1}$

$= \frac{\frac{3 \left(- 8 + 3 x\right) - \left(4 - 3 x\right)}{- 8 + 3 x}}{\frac{4 - 3 x + \left(- 8 + 3 x\right)}{- 8 + 3 x}}$

$= \frac{3 \left(- 8 + 3 x\right) - \left(4 - 3 x\right)}{4 - 3 x + \left(- 8 + 3 x\right)} = \frac{- 28 + 12 x}{- 4}$

$= 7 - 3 x$