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Let $f$ be a one-to-one function whose inverse function is $f^{-1}(x)=x^5+2x^3+3x+1$.

Compute the value of $x_0$ such that $f(x_0)=1$. I am confused as to what this question is asking me, particularly since I don't understand the subscript under the $x$ variable.

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Your first sentence is not a sentence. Format your question with $\LaTeX$. I see neither a variable $x$ nor a subscript. Did you draw that graph? –  Josué May 8 '12 at 5:29
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the subscript variable 'x' means another variable, it is a little bit confusing, you should write it as $f^{-1}(y)=x^5+...$ –  Yimin May 8 '12 at 5:34
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Kurt, the subscript on $x_0$ doesn't really mean much, it's just trying to distinguish $x_0$ from the variable $x$. $x_0$ is a value that satisfies $f(x_0)=1$ –  Alex R. May 8 '12 at 5:35
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I see you have "0% acceptance rate" which is not very nice, please try to accept some of the answers to your questions if they're good enough for that. –  Gigili May 8 '12 at 5:35
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@Yimin: No, writing $f^{-1}(y)=x^5+\dots$ would be absolutely wrong. It's fine as it is. –  Brian M. Scott May 8 '12 at 5:36

3 Answers 3

up vote 2 down vote accepted

$f^{-1}(x)=x^5+2x^3+3x+1$.

$f(f^{-1}(x))=f(x^5+2x^3+3x+1)$.

$f(f^{-1}(x))=x$ as function property enter image description here

$x=f(x^5+2x^3+3x+1)$.

$f(x^5+2x^3+3x+1)=x$

$f(x_0)=1$

You can see that $x=1$ so $x_0=x^5+2x^3+3x+1$

$x_0=1^5+2.1^3+3.1+1=7$

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The point of the question is to see whether you understand how $f$ and $f^{-1}$ are related to each other: $$f\big(f^{-1}(x)\big)=x\;,\tag{1}$$ and $$f^{-1}\big(f(x)\big)=x\tag{2}$$ for any $x$. You want to find an $x_0$ such that $f(x_0)=1$. If $f(x_0)=1$, then $$f^{-1}\big(f(x_0)\big)=f^{-1}(1)\;.$$ Now use $(2)$ the formula for $f^{-1}$ that you've been given, and you'll have your $x_0$.

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Thank you, that greatly clarifies it for me. –  Kurt May 8 '12 at 5:50

One way to find the inverse of an equation is to reverse the variables $x$ and $y$ (this corresponds to reflection over the line $y=x$) and then solve the resulting equation for $y$. In this case, you don't need to actually find the inverse, just compute a specific value. If we have $$ y=x^5+2x^3+3x+1 $$ then we can write the inverse as $$ x=y^5+2y^3+3y+1 $$ Since this second equation is the inverse of the inverse, it is the original equation $f(x)$. We want to find the $x_0$ such that $f(x_0)=1$. By the identification of $y$ with the function, all we have to do to find $x_0$ is plug in $y=1$.

EDIT: Brian's answer is exactly what I've done, only written with better notation. Use that since it makes that relationships much clearer than hiding it behind another variable $y$.

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