# A number a is called a fixed point of a function f if f(a)=a. Consider the function f(x)=x^87+4x+2, x in R. (a) Use the Mean Value Theorem to show that f(x) cannot have more than one fixed point. (b) Use the Intermediate Value Theorem and the result in (a) to show that f(x) has exactly one fixed point.

A number $a$ is called a fixed point of a function $f$ if $f\left(a\right)=a$. Consider the function $f\left(x\right)={x}^{87}+4x+2$, $x\in \mathbb{R}$.
(a) Use the Mean Value Theorem to show that $f\left(x\right)$ cannot have more than one fixed point.
(b) Use the Intermediate Value Theorem and the result in (a) to show that $f\left(x\right)$ has exactly one fixed point.
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Ashlynn Delacruz
(a) If $f$ has two distinc fixed points, namely $a, then
$f\left(b\right)-f\left(a\right)={f}^{\prime }\left(c\right)\left(b-a\right)$
for some $c\in \left(a,b\right)$. Then ${f}^{\prime }\left(c\right)=1$. But ${f}^{\prime }\left(x\right)={x}^{86}+4\ge 4$.
(b)Let $F\left(x\right)=f\left(x\right)-x$, which is continuous. $F\left(0\right)=2$ and $F\left(-1\right)=-2$. So there is $c\in \left(-1,0\right)$ such that $F\left(c\right)=0$ and, hence, $f\left(c\right)=c$. By (a), this is the only possible fixed point.
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gemauert79
$f\left(x\right)-x$ is an increasing function. It is positive at $x=0$ and negative at $x=-1$.