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I'm delving into some further algebra as I prepare for the GRE in 2017!

This practice question is causing me trouble:

The function $f(x)$ is defined by: $f(x) = x^2 - 6x + 13$, $x\in \Bbb R$, $x\ge 3$.
Obtain the inverse function, stating its domain and range.

So first, I'm right in thinking that the inverse function will be the same as the domain of the original function?

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    $\begingroup$ You mean the range of the inverse function will be the same as the domain of the original function. $\endgroup$ – almagest Jun 22 '16 at 11:48
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    $\begingroup$ In general, $\operatorname{Domain}(f) = \operatorname{Range}(f^{-1})$ and $\operatorname{Range}(f) = \operatorname{Domain}(f^{-1})$. $\endgroup$ – user137731 Jun 22 '16 at 11:49
  • $\begingroup$ So far I have switched f(x) and y to give me y = x^2 - 6x + 13 and then I've switched x and y to give me x = y^2 - 6y + 13 $\endgroup$ – New Zealand's finest Jun 22 '16 at 11:55
  • $\begingroup$ Next I will solve for y......which becomes difficult with two y's. Should I have left x^2 as it was? $\endgroup$ – New Zealand's finest Jun 22 '16 at 11:56
  • $\begingroup$ Okay so now I've left x^2 in and have arrived at 6y = x^2 + 13 -x $\endgroup$ – New Zealand's finest Jun 22 '16 at 11:58
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First let's get a bearing on $f$. To do so we'll graph it. The function $$x\mapsto x^2-6x+13$$ looks like this

enter image description here

But the function $f$ has a restricted domain, so let's see which part of this graph remains when we consider only $x\ge 3$.

enter image description here

Notice that this function is strictly increasing (and thus invertible).


Now we'll switch the $x$ and $y$ in this equation and see if we can't find an expression for the inverse:

$$x=y^2-6y+13 \\ x= (y-3)^2-9+13 \\ x=(y-3)^2+4 \\ \pm\sqrt{x-4}=y-3 \\ y=3\pm\sqrt{x-4}$$

enter image description here

Notice that this is not a function because it is multivalued everywhere in its domain except right at $x=4$. So we'll need to choose the correct branch of this function (by taking either the $+$ or $-$ sign in the above equation) given the restriction we had for the domain of $f$.


Now we recall that the inverse of a function looks graphically like the function reflected over the line $y=x$. Confirm for yourself that if you reflect $f$ over that line you get the upper part of the graph of $y=3\pm\sqrt{x-4}$. Thus the inverse function is $f^{-1}(x) = 3\color{red}{+} \sqrt{x-4}$.


Now let's consider the domain and range of both functions. The domain of $f$ is given in the question: $D(f) = \{x\mid x\ge 3\}$. The range is easy enough to get from the graph. Notice the lowest point on the graph is $(3,4)$ and then it increases without bound. So the range is $R(f) = \{y\mid y\ge 4\}$.

The inverse function should be the opposite of this. Meaning the range (domain) of $f$ should be the domain (range) of $f^{-1}$. So let's see. We can see the domain of $f^{-1}$ is $D(f^{-1}) = \{x\mid x\ge 4\}$. It would have been easier to see graphically if WolframAlpha had been more accommodating, but you can see why it must be this -- the $\sqrt{x-4}$ part is undefined for $x<4$. Then, remembering we're only taking the upper branch of the graph above, we see that $f^{-1}$ is an increasing function just like $f$ was. Its lowest point is $(4,3)$ (just plug in $x=4$ to see $y=3$) and it increases without bound from there. So $R(f^{-1}) = \{y\mid y\ge 3\}$ as expected.

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  • $\begingroup$ Thank you so much for taking the time, you managed to crack my mind. I've got it. Super appreciative of you. $\endgroup$ – New Zealand's finest Jun 22 '16 at 12:33
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It is really simple honestly, for this particular equation you can swap $x$ and $y$, and once you get $x=y^2-6y+13$, you solve the equation for $y$, and there you have it, you have the answer.

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  • $\begingroup$ I'm keen to have a go myself, I just don't know how to get the y out of the y^2. Is this where the completion of the square comes in? $\endgroup$ – New Zealand's finest Jun 22 '16 at 12:18
  • $\begingroup$ Which gives (y-3)^2 + 4 $\endgroup$ – New Zealand's finest Jun 22 '16 at 12:19
  • $\begingroup$ You can simply solve it with a Quadratic formula, ax^2+bx+c=0 $\endgroup$ – Alexander Mikky Jun 22 '16 at 12:24
  • $\begingroup$ Oh well, look like somebody did it for ya) $\endgroup$ – Alexander Mikky Jun 22 '16 at 12:27

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