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I'm trying to do this problem but, using the definition I'm given, I cannot get the solution.

I'm asked to show that the equation

$$ f(x) = \cos(x) - x^3 -2x = 0$$

has a unique solution in $[0,1]$. I have the following definition for contractive functions:

Let $g \in C^1[a,b]$ have the following properties

$$g([a,b]) \subset [a,b]$$

$$ |g'(x)| < 1 \hspace{5mm} \forall x\in[a,b] $$

then $g$ is contractive. But the first property doesn't hold here?

$$ f(0) = 1 \hspace{5mm} f(1) \approx-2 $$

What am I doing wrong?

2 Answers 2

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Both $\cos(x)$ and $-(x^3+2x)$ are decreasing functions on $[0,1]$.
It follows that $f(x)=\cos(x)-(x^3+2x)$ is also a decreasing function on $[0,1]$, and since $f(0)=1>0$ while $f(1)<0$, $f(x)$ has a unique root in $[0,1]$. If there were two distinct roots in $[0,1]$, we would have $f'(\xi)=0$ for some $\xi\in(0,1)$, and that is impossible.

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    So the fact that $f(x)$ may or may not be contractive is not important? – 2017-02-01
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    @AnthonyP: no, it isn't. You are not looking for fixed points of $f(x)$, just zeroes. – 2017-02-01
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$f^{\prime}(x)=-\sin(x)-3x^2-2<0$ for all $x \in [0,1]$, so the function is strictly decreasing, and is hence injective. So, there is at most $1$ root, but you can use the Intermediate Value Theorem to deduce the root's existence.

If that property does not hold, it simply means that $f$ is not contractive, you may not be doing anything wrong.