<s>[INST] Numerical analysis - Showing fixed point exists Let $g(x) = \frac{1}{2}(e^{−x})\cos x$. Prove that $g(x)$ converges to a fixed point. 
The answer provided by my lecturer is: 
$g(x)$ is continuous on $[0,1]$, and we can easily verify that $0 ≤ g(x) ≤ 1$ for $x ∈ [0,1]$. Furthermore, since $g'(x) = −\frac{1}{2}(e^{−x})(\sin x + \cos x)$, so $|g'(x)|≤ \sqrt{2}/2 < 1$, for $x ∈ (0,1)$.
My question is: 
Why is $|g'(x)| ≤ \sqrt{2}/2$ ? My understanding is that we are supposed to find the maximum value attainable by $g'(x)$ over the interval $(0,1)$. As such, we should let $g''(x) = 0$, which implies that $\frac{1}{2}(e^{-x})\sin x = 0$ (but we get $x = 0$ which is out of the interval $(0,1)$.
Can anyone point out any misconceptions I have?  
 [/INST] The estimate of $|g'(x)| \le \sqrt{2}/2$ can be justified by combining $0 \lt e^{-x} \le 1$ on the interval $[0,1]$ and the bound $|\sin x + \cos x| \le \sqrt{2}$:
$$ |g'(x)| = \frac{1}{2} e^{-x} |\sin x + \cos x| \le \frac{\sqrt{2}}{2} $$
The estimate of $e^{-x}$ is easy because that is a monotone decreasing function whose maximum is its value at $x=0$, namely $1$.
The maximum of $|\sin x + \cos x|$ is quickly found by a standard calculus argument, or by using a trigonometric identity:
$$ \sin\left( x + \frac{\pi}{4} \right) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} (\sin x + \cos x) $$
$$ \sin x + \cos x = \sqrt{2} \sin\left( x + \frac{\pi}{4} \right) $$
$$ |\sin x + \cos x| \le \sqrt{2} $$
Since the above is true globally (for all real $x$), it holds on $[0,1]$.
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