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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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    @BrianM.Scott I see. Thx.2012-05-08

3 Answers 3

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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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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.2012-05-08
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$f^{-1}(x)=x^5+2x^3+3x+1$.

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

$f(f^{-1}(x))=x$ 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 $x_0=x^5+2x^3+3x+1$

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