Given function $f(x)$ is $1-\sqrt{4-2x}$ and $g(x)$ is $3-2x$, find a function $h(x)$ where $h(g(x)) = f(x)$

Question

My steps:

$h(3-2x(x)) = 1-\sqrt{4-2x}$

$1-\sqrt{4-2(3-2x)}$

$1-\sqrt{4-6+4x}$

$1-\sqrt{-2+4x}$

But apparently, the correct answer is $1-\sqrt{1+x}$. Can you please explain?

Answer 1

Let $y:=3-2x$, then $2x=(3-y)$. You want $h(3-2x)=1-\sqrt{4-2x}$, i.e. $h(y)=1-\sqrt{4-(3-y)}= 1-\sqrt{1+y}$.

Answer 2

$4-2x=1+(3-2x)$, so

$$h(g(x)) = f(x) \iff h(3-2x) = 1 - \sqrt{1+(3-2x)} \implies h(x) = 1-\sqrt{1+x}$$

Answer 3

Let

$$f(x) = 1-\sqrt{4-2x}$$ $$g(x) = 3-2x$$

You want to find a function $h(x)$ such

$$h(g(x)) = f(x)\label{1}\tag{1}$$

As we know, the inverse of a function applied on the same function is $x$. Therefore:

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

Putting $g^{-1}$ as $x$ in \eqref{1} you get

$$h\left(\underbrace{g\left(g^{-1}\right)}_{x}\right) = f\left(g^{-1}\right)$$ $$h(x) = f\left(g^{-1}\right) $$ $$h(x) = 1-\sqrt{4-2\cdot g^{-1}}\tag{2}\label{2}$$

Once $g(x) = 3-2x$, then

$$g^{-1}(x) = \dfrac{3-x}{2}\label{3}\tag{3}$$

Then putting \eqref{3} into \eqref{2} we get

$$\begin{align*}h(x) & = 1-\sqrt{4-2\cdot \left(\dfrac{3-x}{2}\right)} \\ & = 1-\sqrt{1+x}\end{align*}$$

Answer 4

The functions are $f(x) = 1 - \sqrt{4 - 2x}$ and $g(x) = 3 - 2x$.

Since $g(x)$ is a linear function, it is also invertible. So we can find $g^{-1}(x)$ like this: $$g(g^{-1}(x)) = 3 - 2\cdot g^{-1}(x)$$ $$\implies x = 3 - 2\cdot g^{-1}(x)$$ $$\implies g^{-1}(x) = \frac{3-x}{2}$$

If $\ h(g(x)) = f(x)$, $\ \therefore h(g(g^{-1}(x))) = f(g^{-1}(x))$ $$\implies h(x) = 1 - \sqrt{4 - 2\left(\frac{3-x}{2}\right)}$$ $$\implies h(x) = 1 - \sqrt{1+x}$$

Answer 5

You need to work out how to get $3-2x$ to $1-\sqrt{4-2x}$ by operations. You first need to add $1$ and then take the square root to get $\sqrt{4-2x}$. Then you need to multiply by $-1$ and add $1$ to get $-\sqrt{4-2x}+1$, which is $1-\sqrt{4-2x}$. To represent this series of operations as an equation, you start with $x$. You then add $1$, take the square root, multiply by $-1$ and add $1$ again to get $-\sqrt{x+1}+1$, which is $1-\sqrt{x+1}$.

The mistake you made is treating the $g$ of the $g(x)$ as a variable. $g(x)$ does not equal $3-2x(x)$, it actually equals $3-2x$.