Let $f:[0,1] \to \mathbb{R}$ defined by $$ f(x)=\begin{cases} \frac{1-\cos x}{x}, &x \ne 0, \\ 0, &x=0. \end{cases} $$ I want to find the maximum value of $f$.
The derivative $f'(x) = \frac{x \sin x + \cos x - 1}{x^2}$ doesn't give me much information. It seems that its graph is increasing on $[0,1]$, so I think its maximum occurs at $x=1$.
However, is there a rigorous way to find the maximum value of the function?
Thank you.
Continuing where you left off, let $g(x)=x\sin(x)+\cos(x)-1$.
We will show that $g(x)$ is positive, for any $x\in [0,1]$.
$g'(x)=x\cos(x)$, which is equal to $0$ for $x=0$ and strictly greater than $0$ for $x\in (0,1]$.
Therefore, $g(x)$ is strictly increasing on the interval $[0,1]$, so $g(x) > g(0)=0$, for $x \in [0,1]$.
But, $f'(x)=\frac{g(x)}{x^2} > 0$, for $x\in[0,1]$.
So $f$ is also strictly increasing on the interval $[0,1]$, which means that the maximum is $f(1)\approx 0.46$.