Evaluate $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$

Question

If $g\left( x \right) = ax + c$, $f\left( x \right) = {x^b} + 3$, and $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$ what is the value of $a+b+c$?

My approach is as follows

Given $g\left( x \right) = ax + c$ & $f\left( x \right) = {x^b} + 3 \Rightarrow {\left( {f\left( x \right) - 3} \right)^{\frac{1}{b}}} = x$

$\frac{{g\left( x \right) - c}}{a} = x$

${\left( {f\left( x \right) - 3} \right)^{\frac{1}{b}}} = \frac{{g\left( x \right) - c}}{a}$

$g\left( x \right) = g\left( x \right) \Rightarrow {g^{ - 1}}\left( {g\left( x \right)} \right) = x$

I am not able to proceed.

Answer 1

Notice that $$(g\circ f)(x) = 2x^3+7$$

But by definition of $g$ and $f$ we have also $$(g\circ f)(x) =ax^b+3a+c$$

Now if $x=0$ then we get $7 =3a+c$ and then $ax^b = 2x^3$. So if $x=1$ we get $a=2$ and $c=1$ and $x^b=x^3$. So if $x=2$ we get $2^b = 8$ and thus $b=3$.

Notice: Since we don't know apriori that $b$ is positive integer you can not do just the comparison of the coefficients.

Answer 2

Hint: Recall that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$. You've already compute each inverse separately, so now we just compose them to get: $$ (g \circ f)^{-1}(x) = f^{-1}(g^{-1}(x)) = f^{-1}\left(\frac{x - c}{a}\right) = \left( \left(\frac{x - c}{a}\right) - 3 \right)^{\frac{1}{b}} = \left( \frac{x - 3a - c}{a}\right)^{\frac{1}{b}} $$

Answer 3

You can start by noting that $$ (g \circ f)(x) = a(x^b+3)+c $$

and, inverting this expression, get to $$ (g\circ f)^{-1}(x) = \left( \frac{x-c-3a}{a} \right)^{\frac 13} $$

Identifying coefficients, you'll see that $a=2, b=3, c=1$, and consequentely, $a+b+c = 6$.