How do I find $f(x)$ based on this?

Question

So I have this equation:

$$f\left(\frac{x}{3-2x}\right)= x^3-5,$$ and I am supposed to find $f\left(-\frac{2}{7}\right).$ I started by trying to find $f(x)$ based on what was given, but I think I completely confused myself, as I tried to multiply the $x^3-5$ by $3- 2x$ and other stuff like that, and then tried to plug in $f\left(\frac{x}{3-2x}\right)$ back into what I got. Can you please help me at least find $f(x),$ or a simpler way to get $f\left(-\frac{2}{7}\right)?$

Thanks a bunch!

Answer 1

If you let $$y=\frac{x}{3-2x},$$ then what do you get $f(y)$ in terms of $y$?

Answer 2

Guide:

Let $$\frac{x}{3-2x}=\frac{-2}7$$

Solve for $x$ by converting it to a linear equation.

Now compute $x^3-5$.

Answer 3

HINT

• from $\frac{x}{3-2x}=-\frac27 \implies 7x=-6+4x\implies3x=-6$ then find $x$
• evaluate $x^3-5$

the answer should be

-13

Answer 4

$(f\circ g)(x) = x^3 - 5\\ g(x) = \frac {x}{3-2x}$

We haven't found $f(x)$ yet, but probably don't need to.

We need to find an $x$ such that $g(x) = -\frac {2}{7}$

"By inspection," it just seems kind of obvious that $x = -2$

But if that didn't just jump out at you.

$-\frac {2}7 = \frac {x}{3-2x}\\ 4x-6 = 7x\\ 3x = -6\\ x = -2$

$(f\circ g)(-2) = -8 - 5 = -13$