Find inverse of $7x^{2}-112x+448$

Question

Given the function $\; f(x) = 7x^{2}-112x+448, \;$ for $x\ge 8, \;$ find $\displaystyle \;$ $f^{-1}(x)$.

To find inverse, I should just solve for x in terms of y:

$$y = 7x^{2}-112x+448$$

I can rewrite it as:

$$y = (7x-56) (x-8)$$

but what's next?

Or I can also say:

$$y - 448 = 7x (x - 16)$$

and it gets me nowhere ...

p.s. I don't care about signs (e.g., $\; x -8 >= 0$) for now, cuz that's not the hard part

Answer 1

Note that you can rewrite $7x^2-112x+448=0$ as $7(x-8)^2=0$

Answer 2

Following the comments and evinda's answer, you write $y = 7x^2-112x+448$ as $y = 7(x-8)^2$. So to write $x$ as a function of $y$, we take roots and obtain $\sqrt{y} = \sqrt{7} |x-8|$. Without the condition $x \geq 8$ we can't get a single value for $y$ (the function isn't injective in $\Bbb R$). The condition $x \geq 8$ allows us to write $|x-8| = x-8$, and now you can finish it.

Answer 3

I think that what you're missing a a method called "Completing The Square". Note what happens when we factor out the leading coefficient $$ax^2+bx+c = a(x^2+\frac{b}{a}x+\frac{c}{a})$$ Now, by multiplying out $(x+t)(x+t)=x^2+2tx+t^2$ we can see that in a perfect square that $\frac{1}{2}$ of the middle coefficient, squared, is equal to the last [constant] coefficient. Applying this model to our original expression we have $$\begin{array}{lll} a(x^2+\frac{b}{a}x+\frac{c}{a}) &=& a\bigg(x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2\bigg)\\ &=& a\bigg((x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a^2}\bigg)\\ \end{array}$$ In the case where $b^2-4ac=0$, this expression will reduce to $$a(x-\frac{b}{2a})^2$$

Lets follow these steps with $$y=7x^2-112x+448$$ First, let's factor $7$ from each term $$y=7(x^2-16x+64)$$ By noting that $\bigg(\frac{1}{2}\cdot (-16)\bigg)^2 = (-8)^2=64$, we know that we already have a perfect square within the parentheses. Half of the middle term ($-16$) is $-8$, so it factors as follows $$y=7(x-8)^2$$ I hope this helps.

Answer 4

You need to solve the equation: $$ 7x^{2}-112x+(448-y)=0 $$ The roots are $$ \frac{112+\sqrt{112^2-4\cdot 7(448-y)}}{2\cdot 7} =8+\frac{\sqrt{7y}}{7} $$ and the companion $$ \frac{112-\sqrt{112^2-4\cdot 7(448-y)}}{2\cdot 7} =8-\frac{\sqrt{7y}}{7} $$ Since you know that $x\ge8$, only the first root can be used and you get also that the domain of the inverse is $y\ge0$.