What is the maximum of $\frac{(1-\cos x)}{x}$ in the interval $[0, \pi]$? What is the maximum of ${\frac{(1-\cos x)}{x}}$ in the interval $[0, \pi]$?
I can show that the maximum is less than 1, but I want an exact value.
 A: Find the critical values. Since it is a closed interval, two of them will be $x=0$ and $x=\pi$. The others will be points where $\frac{d}{dx}\frac{1-\cos{x}}{x}=0$. Evaluate the function at each critical point and observe which one is/which ones are the smallest.
A: I do not know the answer analytically, but I suggest a numerical solution. It's pretty easy to find such things online; in the interest of homework rules, report back if you try and still can't find the answer.
A: By the quotient rule, the derivative of your function is: 
$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1-cos(x)}{x} \right )=\frac{x\sin x+\cos x-1}{x^2}$
The maximum will occur when this is equal to 0: 
$\frac{x\sin x+\cos x-1}{x^2}=0$
which yields, by multiplying through by $x^2$:
${x\sin x+\cos x-1}=0$
We are looking for the solutions of this equation in the interval $[0,\pi]$. I do not believe that this equation can be directly solved by using general methods (trig identities, etc). The solution in the designated range is approximately $2.33$, through wolfram alpha.
I cannot provide a method of calculating it, only because there does not seem to be a "simple and clean" way to solve the equation. 
In order to ensure that this is our maximum, we must calculate the second derivative of $f(x)$, and then calculate $f''(2.33)$ to ensure that the value is indeed negative. If the value is negative, then $f(x)$ is concave down at $x=2.33$, and therefore, we have a maximum. 
$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{x\sin(x)+\cos (x)-1}{x^2} \right)=\frac{\cos (x)(x^2-2)-2x\sin (x)+2}{x^3}$
which can be found by using the quotient rule once again. Evaluating at $2.33$ yields $f''(2.33)=-0.3$ (approximately)
Since this is negative, we know that we have found our maximum. 
Thus, the maximum value occurs at $f(2.33)$. We can evaluate this to be $0.724611...$. However, because of the "ugly" nature of the derivative I was not able to give a closed form solution. Some things just can't be solved by hand.
