Find the inverse function of $g(x)=(x-2)(x-4),\; x≥3$.

Question

Find the inverse of the following function, stating its domain.

$$ g(x) = (x-2)(x-4), \quad x≥3. $$

I try to find the inverse function, but I can't eliminate $x$ in my method.

Answer 1

Hint: $0=x^2-6x+(8-y)$ Know a formula to find $x$ now?

Answer 2

The "switch and solve method".

$g(x) = y = (x - 2)(x - 4) = x^2 - 6x + 8$. Swapping $x$ and $y$, we have:

$x = y^2 - 6y + 8$, we try to solve for $y$.

$x = y^2 - 6y + 9 - 1$

$x + 1 = (y - 3)^2$

$y - 3 = \pm \sqrt{x + 1}$

$y = \pm \sqrt{x + 1} + 3$

Now pick the proper sign for the square root, based on $x \geq 3$, and find the domain.

Answer 3

$(x-3)^2=g(x)+1\iff x=\pm\sqrt{g(x)+1}+3$

Since $x\ge 3$, necessarily $x=\sqrt{g(x)+1}+3\iff g^{-1}(y)=\sqrt{y+1}+3$.

To find domain, $y+1\ge 0$.

Answer 4

First, rewrite $y = g(x)$ in vertex form and solve for $x$: $$ \begin{align} y &= (x - 3)^2 - 1 \\ y + 1 &= (x - 3)^2 \\ \sqrt{y + 1} &= \lvert x - 3 \rvert \\ \sqrt{y + 1} &= x - 3 \tag{since $x \ge 3$} \\ 3 + \sqrt{y + 1} & = x \end{align} $$

Writing this inverse as a function: $g^{-1}(y) = 3 + \sqrt{y + 1}$, or if you switch variables, $$ g^{-1}(x) = 3 + \sqrt{x + 1}. $$

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Here's a sketch of the graphs $y=g(x)$ and $y=g^{-1}(x)$ on the same axes (with the dashed line $y=x$ to illustrate the reflection symmetry).

Answer 5

Given $g(x)=(x-2)(x-4)$, Observe, when $x\to x+3$

$$g(x+3)=(x+1)(x-1)=x^2-1=[(x+3)-3]^2-1$$

So for $x+3\to x$

$$y=g(x)=(x-3)^2-1$$

As $x\ge3$,

$$y+1=(x-3)^2$$ $$g^{-1}(y)=x=3+\sqrt{y+1}$$