# Functions and Algebraic Operations

I just need some help on these equations:

$f: A \to B$ and $y=f(x)=2x^{3}+1\implies f^{-1}(x)=$ ?

THANK YOU very much for reading this and please, pardon me that I'm STILL not skilled enough to use proper or universal mathematical symbols on the Web those might have been recognized almost all around the World...

I just checked the equation and please let me know if it's not ' y=2x^(3)+1 '.

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Could someone show me how to calculate this problem step by step, please... Thank you... – Kerim Atasoy Feb 28 '12 at 18:57
Do you want $y=2x^3+1$? – David Mitra Feb 28 '12 at 19:02
Oh, sorry, yes, it's \' y=2x^(3)+1\'\. I guess it's done now...? – Kerim Atasoy Feb 28 '12 at 19:25

If you mean $y = f(x) = 2x^3 + 1$, then $f$ does have an inverse. You can calculate it as follows: $$y = 2x^3 + 1 \Rightarrow y - 1 = 2x^3 \Rightarrow \frac{y-1}{2} = x^3 \Rightarrow \sqrt[3]{\frac{y-1}{2}} = x,$$ so to get from a value of $y$ you apply $$f^{-1} = \sqrt[3]{\frac{x-1}{2}}$$

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-1 for saying that $\sqrt[4]{\frac{x}{2}}$ is not a function – TMM Feb 28 '12 at 19:18
Thank you very much. It seems like the lecturer made some mistake, too... :) "Nobody is perfect", right... :) – Kerim Atasoy Feb 28 '12 at 19:21

Recall the definition of the inverse function $f^{-1}$. $f^{-1}(y)$ is a number; it is the input value $x$ of $f$ such that $f(x)=y$. In other words, $f^{-1}(y)$ is the input value for the function $f$ that produces the output value $y$.

In functional notation: $$f(x)=y,\quad \text{ if and only if }\quad f^{-1}(y)=x.$$

Let's look at your function, which I assume to be $f(x)=x^3+1$. If you set $y=f(x)$, then you have $$\tag{1} y=x^3+1$$

In the above, $x$ is the input value for $f$ that gives the output value $y$. Equation $(1)$ gives a formula for determining the output value $y$ of $f$ when you know what the input value $x$ is. Equation $(1)$ is the rule for the function $f$. For instance if the input were $x=1$, the output would be $y=1^3+1=2$. In functional notation, you would write $f(1)=2$.

The inverse function sort of "reverses things". You specify an output value of $f$, such as $y=2$, and the inverse function tells you what the input to $f$ was. For example, if the output of $f$ was 2, then the input had to have been $1$. In functional notation $f^{-1}(2)=1$.

It's worth repeating: in general if $f(x)=y$, then $x=f^{-1}(y)$.

What you need to do in your problem is to find the rule for $f^{-1}$. That is, you need to find a general formula that tells you how to compute the input $x$ if you know what the output $y$ is. Well, we almost have one in hand: we just need to solve equation $(1)$ for $x$ in terms of $y$.

That is, we need to isolate $x$ on one side of equation $(1)$ and insure that there are "no $x$'s" on the other side. Then on one side $x$ will be by itself, and the other side of the equation will be the rule for $f^{-1}(y)$.

Let's do this: Start with equation ${1}$. $$y=x^3+1.$$ We want to get $x$ by itself on one side of the equation and no $x$'s on the other. We may (and do) subtract 1 from both sides of the above equation. This gives: $$y-1=x^3.$$ We don't want $x^3$ on the right hand side, we just want $x$. Take the cube root of both sides to obtain: $$\root 3\of{ y-1}=x.$$ And just to make things pretty, write the above equation in the opposite order: $$\tag{2} x = \root 3\of{ y-1}$$ This is what we want. If we know $y$, then we can use equation $(2)$ to figure out what $x$ was.

For instance, if $y=28$, what was the input value $x$ for $f$? The formula above tells you: $x=f^{-1}(28)=\root 3\of{28-1}=3$.

More generally formula $(2)$ tells you that $$f^{-1}(y) = \root 3\of{ y-1}.$$ In more conventional notation (that is, using $x$ to denote the independent variable): $$f^{-1}(x) = \root 3\of{ x-1}.$$ and we're done.

More generally:

Keep in mind that not all functions have inverses. Only one-to-one functions do.

If $f$ is one-to-one it has an inverse. To find the rule for $f^{-1}(x)$, you can do the following:

1. Write down the equation $y=f(x)$ (where $f(x)$ is replaced by the rule you were given ($x^3-1$ in your example).
2. Solve the equation for $x$. You want an equation of the form $x=\text{stuff with only } y{\text{'s}}$.
3. Write down $f^{-1}(y) = \text{stuff with only } y{\text{'s}}$.

4. Replace in the formula all $y$'s with $x$'s.

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:) "AW(E)SOME!"... :) THANK YOU, sir... – Kerim Atasoy Feb 28 '12 at 20:26
Kerim, I would change this to the accepted answer. The other still contains an error (typo at best, conceptual misunderstanding at worst) in the line "to get from a value of y..." – The Chaz 2.0 Mar 6 '12 at 4:57
Well, thank you very much for noticing me and all others but, actually I don't care about "reputation" around these sites so much... :) I'm here, bacause, I really would like to learn something useful and that might work for me as well. By the way, I mean, by doing this kind of things here, I'm having a nice chance to meet new people or friends which might be "a little bit different" from other people, see... :) They seem really experienced sometimes and some of them might be very brightful also. I REALY respect intelligence which is used for good... – Kerim Atasoy Mar 6 '12 at 9:34
You people have been helping me so much actually... – Kerim Atasoy Mar 6 '12 at 9:34