# Finding the inverse function of a quadratic function [closed]

let $$f:[-4,∞) \rightarrow \mathbb{R} , f(x)=-(x+4)^2 +3$$. show that $$f^{-1}:(-∞,3] \rightarrow \mathbb{R}, f^{-1}(x)=\sqrt{3-x}-4.$$

## closed as off-topic by Eevee Trainer, Gregory J. Puleo, Nosrati, Shailesh, Martin ArgeramiMay 8 at 3:45

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• Please use Mathjax. – Abstract Analysis May 8 at 1:52
• It's the inverse function, not the reciprocal. – Robert Israel May 8 at 2:08
• @LeAnhDung will use in future, thank you – chloe rogers May 8 at 2:16

Just follow this procedure when trying to calculate inverse functions:

STEP 1:

Write the initial function

$$f(x) = -(x+4)^2+3$$

as

$$y = -(x+4)^2+3$$

STEP 2:

Replace $$y$$ with $$x$$ like so

$$x=-(y+4)^2+3$$

STEP 3:

Solve the above in order to separate $$y$$

$$\sqrt{3-x} -4 = y$$

STEP 4:

Replace this new $$y$$ with $$f^{-1}(x)$$

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

Find the domain of this new function. Can you proceed?

STEP 5:

Check your work knowing that $$f^{-1}(f(x))=x$$

If you do some algebra, verify that your result (in this case)

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

• Good and clear. One more step, is to prove that the inverse function found is indeed the correct function (optional). – NoChance May 8 at 2:07
• thank you. As for the domain restriction, that is only found by graphing the inverse function? – chloe rogers May 8 at 2:25
• @chloerogers you can find the domain by basically identifying what values of $x$ are not allowed (because of the square root in this case). so in this case, $x$ cannot be greater than 3 for the function to be real – Dashi May 8 at 2:27

Hint:

For $$x\ge-4$$, solve the equation $$-(x+4)^2+3=y$$ and find $$x$$ in term of $$y$$.