If $f(g(x)) = 4x^2-8x$ and $f(x)=x^2-4$, then what's the value of $g(x)$?

Question

I'm a little stuck with this simple function:

If ${f(g(x)) = 4x^2-8x}$ and ${f(x)=x^2-4}$, then what's the value of ${g(x)?}$

Any tips?

Answer 1

By setting $y:=g(x)$ we see that: $$ 4x^2-8x\,=\,f(g(x))\,=\,f(y)\,=\,y^2-4\,=\,g(x)^2-4 $$

Thus: $$ g(x)^2 = 4(x^2-2x+1)=4(x-1)^2 $$

which implies that $g(x)=\pm2(x-1)$.

Answer 2

Hint: complete the square in $4 x^2 - 8 x$, to express it as $(\ldots)^2 - 4$.

Answer 3

If $g$ is a polynomial, then note that the degree can't be higher than $1$. Suppose $g(x)=ax+b$. Then

$$f(g(x))=f(ax+b)=a^2x^2+2abx+b^2-4$$

Now we want $a^2=4$, $2ab=-8$, and $b^2-4=0$. Two such solutions are $a=2,b=-2$, or $a=-2,b=2$.

Pluggin those back in, we have

$$f(2x-2)=4x^2-8x+4-4=4x^2-8x$$

and

$$f(-2x+2)=4x^2-8x+4-4=4x^2-8x.$$

Answer 4

Since $4x^2-8x=g^2-4$, $g^2=\left( 2x-2\right)^2$ so $g$ may be chosen as either $2x-2$ or $2-2x$, or even a function that can be either depending on the value of $x$, e.g. $\left| 2x-2\right|$ or $-\left| 2x-2\right|$.

Answer 5

\begin{align} f(g(x))&=4x^2-8x\\ g(x)^2-4&=4x^2-8x\\ g(x)^2&=4x^2-8x+4\\ &=4(x^2-2x+1)\\ &=4(x-1)^2\\ &=[\ \!2(x-1)\ \!]^2\\ \implies g(x)&=\pm2(x-1) \end{align}