# What is the Inverse function of $f(x)=192x-16x^2$?

What is the inverse function of: $$f(x)=192x-16x^{2}$$

I have been finding myself going in a circle in trying to complete this problem which otherwise looks simple, but for some reason I am at a block. Could someone help explaining how to go about this problem?

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This function is not monotone since $f'(12)=0$ hence it is not $1-1$ thus not invertible – Belgi Sep 2 '12 at 19:14

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@Belgi Even though your four pieces of fact are correct, the "hence"-part is wrong. There are loads of non-monotone functions which are 1-1 and perfectly invertible, even assuming everywhere-differentiability. $x^3$ is the easiest example. – Arthur Sep 2 '12 at 19:26
@Arthur $x^3$ is monotone. What it fails is having a nonvanishing derivative. If you want a non-monotonic injective function, by IVT it will have to be discontinuous. – Logan Maingi Sep 2 '12 at 19:31
@Arthur: $f(x)=x^3$ is monotone. It’s even strictly monotone. – Brian M. Scott Sep 2 '12 at 19:32
I'm sorry, it's the "since"-part that is wrong, then. – Arthur Sep 2 '12 at 19:36
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## 4 Answers

Instead of completing the square, as in lab bhattacharjee’s answer, you can use the quadratic formula. You have $y=192x-16x^{2}$; rewrite it as $16x^2-192x+y=0$, and treat $y$ as the constant term to get

$$x=\frac{192\pm\sqrt{192^2-64y}}{32}=6\pm\frac{\sqrt{64(576-y)}}{32}=6\pm\frac{\sqrt{576-y}}4\;.$$

This gives you two functions, both defined for $y\le 576$:

$$x=6+\frac{\sqrt{576-y}}4\;,$$ and $$x=6-\frac{\sqrt{576-y}}4\;.$$

The first corresponds to the righthand side of the parabola $y=192x-16x^2$, and the second to its lefthand side.

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The answers of @nayrb and @lab bhatacharjee are correct as far as they go, but as a (retired) teacher, I feel that I must address a fault of the math education system, here in the States as well as, perhaps, elsewhere. To describe a function, you need to mention the domain and the target space, and that is particularly important here. Sketch the graph! You see that there’s a maximum at $(6,576)$, and that on either side of the vertical line $x=6$, there are points on the graph at equal height. So the function fails the “horizontal line test” unless you restrict the domain. Let’s restrict to the interval $\langle-\infty,6]$, i.e. the closed half-line to the left of $6$. But what about our inverse function? It’s not defined for $y>576$, and that means that its domain has to be no bigger than $\langle-\infty,576]$. It’s only now, once we’ve restricted both the domain and the target space of our original function, that we really have an inverse function, and its formula is indeed the one given by @lab, with the minus sign of course.

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Hi Jonathan! This originated in my calculus class here at UConn. Thanks for the consultation. – Jeremy Teitelbaum Sep 2 '12 at 19:43
A function is an object with three parts: a domain, a range (allowable set out outputs) and a rule tying each element of the domain to some element of the range.  Failure to specify these is failure to completely specify a function.  If you are asking if a function is 1-1 or onto, this is especially important. – ncmathsadist Sep 2 '12 at 20:18
@ncmathsadist, thanks for the confirmation. Only thing is, in high-schools nowadays, it seems that the word “range” is used for what I learned to call the image, and no word at all is used for what I have just called the target space. So I think that the word “range” can be misunderstood by many. – Lubin Sep 3 '12 at 20:49
People fail to make this important distinction.  The range is the set of allowable outputs. The image is the set of realized outputs.  When the image and the range coincide, the function is onto.  I have taught HS math and computer science.  I always have a discussion about the importance of specifying domain and range in functions in math and CS classes. We call them "pure functions" in a CS class. – ncmathsadist Sep 3 '12 at 20:51
Sorry to rejoin this discussion late. @ncmathsadist, you and I are on the same side, but I don’t think we have seen the same phenomena. In the high-school where I volunteer, the “range” is the set of values taken on. And I have to say that more than 50 years ago, George Mackey, a far greater mathematician then either of us, routinely said in class, “Let $f$ be a function with domain $S$, and with range in $\mathbb{R}$.” To me, this says clearly that for him, the range was the set of values taken on. For this reason, I sedulously avoid the use of the word. – Lubin Sep 5 '12 at 0:59
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Let $f(x)=y=192x-16x^2\implies -y=16(x^2-12x)=16(x-6)^2-16\cdot 36$ $\implies 576-y=16(x-6)^2$

$\implies x=f^{-1}(y)=6±\frac{\sqrt{576-y}}{4}$

Clearly $y ≤576$ to make $f(y)$ invertible in real numbers.

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you lost me at 16(x-6)^2. where does that come from? I appreciate the help – Carl Strum Sep 2 '12 at 19:20
@CarlStrum $-y=16(x^2-12x)= 16(x^2 -12x +36 -36) = 16((x-6)^2 -36) =16(x-6)^2-16\cdot 36$ – Arthur Sep 2 '12 at 19:22
Do you think you could use words to explain what you just did? How did you just know to choose 36 to add/subtract to that equation. (i mean i know it doesn't change the problem, but how did you know to use the specific number 36?) – Carl Strum Sep 2 '12 at 19:27
$x^2-12x=(x)^2-2(x)(6)+(6)^2-(6)^2=(x-6)^2-36$, is it ok? – lab bhattacharjee Sep 2 '12 at 19:29
I appreciate all the help but I feel like I'm being misunderstood.  I am interested in how you knew to use whatever technique it was to get those answers.  I.E. I am a little behind on the concept of "completing the square" if that is what's being used to solve the problem. – Carl Strum Sep 2 '12 at 19:33
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The function will only have an inverse if you restrict it to a region where it is one-to-one. You can look at this graph:

and see that you need to restrict your function to one of the regions where $6\le x\le \infty$ and $-\infty<y\le 576$ or $-\infty<x\le 6$ and the same $y$-region.

The quadratic formula (as indicated above) can give you the proper formula. Wolfram Alpha works it out nicely here: Wolfram Alpha Knows the Quadratic Formula You can "show steps" to see how Wolfram Alpha does this; it's essentially completing the square.

The two solutions correspond to the two branches of the inverse function.

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Looks like you’re relatively new here. Welcome! - jdL – Lubin Sep 2 '12 at 19:47
I'm trying to see if I can get my Calc students to use this site as a resource. – Jeremy Teitelbaum Sep 2 '12 at 19:51
It’s more than a resource, ’cause they’ll see mathematics at many different levels. Should be an eye-opener! – Lubin Sep 2 '12 at 19:52
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