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Given the function $$ f(x)=\frac{1}{4}\left((x-1)^{2}+7\right) $$

The first part of the question asks to find the largest domain containing the value $x=3$ for which $f^{-1}(x)$ exists. I determined the domain to be $x≥1$.

The second part of the question is:

Let $a$ be a real number not in the domain found in the previous part, find the exact value of $f^{-1}(f(a))$.

My thinking process was since $a<1$, based from the domain we found previously, then therefore $f(a)=f(-a)$.

Do I use the inverse function i.e. $f^{-1}(x)=1+\sqrt{4x+7}$ and just sub in $-a$? I'm not entirely sure if this is correct. Any help is greatly appreciated!

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  • $\begingroup$ I just had a thought, instead of immediating subbing in $-a$, do I sub in $ f(-a)=\frac{1}{4}\left((-a-1)^{2}+7\right) $ into $f^{-1}(x)$? $\endgroup$ – spuddy Jul 31 '20 at 22:47
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    $\begingroup$ Are you sure about $f(a) = f(-a)$? The function is not symmetric, but it is symmetric around $1$, which means $f(1+x) = f(1-x)$. $\endgroup$ – Raoul Jul 31 '20 at 22:51
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If the function were symmetric about $x=0$, then you would have $f(-x) = f(x)$. In your case, it's symmetric about $x=1$, so the relationship is instead $f(1-x) = f(1+x)$. Thus $$f^{-1}(f(1-x)) = f^{-1}(f(1+x)) = 1+x.$$

Now substitute $a=1-x$.

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  • $\begingroup$ just for clarification, does this mean the "exact value of $f^{-1}(f(x))$ would just be $2-a$? $\endgroup$ – spuddy Jul 31 '20 at 22:55
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    $\begingroup$ Yes, that's right. (I think you mean $f^{-1}(f(a))$.) You can try substituting the actual inverse function if you want some practice with that, but conceptually, it isn't necessary to work out those details. The relationship above will be true for all functions symmetric about $x=1$. $\endgroup$ – Théophile Jul 31 '20 at 22:58
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    $\begingroup$ Side comment: I think the goal of the question might be slightly confusing because of the wording "the exact value". The question is really just "find $f^{-1}(f(a))$". $\endgroup$ – Théophile Jul 31 '20 at 23:05

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