Fixed Points of Polynomial (Application of Mean Value and Intermediate Value Theorems)

Question

The question is:

A number $a$ is called a fixed point of a function $f$ if $f(a)=a$. Consider the function $f(x)=x^{87}+4x+2, x\in\Bbb R.$

(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.

I went to my prof for help, but all he said is that you can solve (a) using a proof by contradiction, but I'm not sure how to do that or what he meant? Any help would be appreciated in explaining! I've gotten as far as letting $f(b)=b$ and figuring out that $f'(c)=1$ but I do not understand the meaning behind it. Thanks in advance :)

Answer 1

• Suppose by contradiction there exist $a<b$ such that $f(a)=a$ and $f(b)=b$. Define the function $g\colon x\in\mathbb{R} \mapsto f(x) -x = x^{87}+3x +2$, so that $g(a)=g(b)=0$. What happens when you apply the MVT to $g$ on $[a,b]$?

  Note that $g^\prime(x) = 87x^{86} + 3 > 0$ for all $x$.

• To apply the IVT, now, observe that $g$ has odd degree ($87$); in particular, $\lim_{-\infty} g = -\infty$ and $\lim_{+\infty} g = +\infty$. How can you use the IVT to argue that there must be some $x$ such that $g(x)=0$?