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

a question from my 11th-grade maths assignment. I don't even know where to start. please help.

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

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

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

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    $\begingroup$ Good and clear. One more step, is to prove that the inverse function found is indeed the correct function (optional). $\endgroup$ – NoChance May 8 at 2:07
  • $\begingroup$ thank you. As for the domain restriction, that is only found by graphing the inverse function? $\endgroup$ – chloe rogers May 8 at 2:25
  • $\begingroup$ @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 $\endgroup$ – Dashi May 8 at 2:27
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Hint:

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

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