How to prove non negativeness of $f(x)=1-\cos x-\frac{x^2}{20}$ I'm trying to prove that $f(x)=1-\cos x -\frac{x^2}{20}$, defined on $[-\pi, \pi]$, is a non negative function.
How do I prove that $f(x)\ge 0 $ for all $x \in [-\pi, \pi]$?
Hints?
 A: A strict, calculus-student friendly approach :
Starting off, one can see that $f(x)$ is an even function, because : 
$$f(-x)= 1 - \cos(-x) - \frac{(-x)^2}{10} = 1 - \cos(x) - \frac{x^2}{10}= f(x)$$
This means that you just have to study the interval $[0,\pi]$.
Taking the derivative of $f(x)$, we have  :
$$f'(x) = \sin(x) - \frac{x}{10}$$
It is $f'(x) > 0 \quad \forall  \quad x\in [0,\pi/2]$ , thus meaning that $f(x)$ is increasing on the interval $[0,\pi/2]$. 
Now, going over the interval $[\pi/2, \pi]$, let's take the second derivative to help ourselves a bit : 
$$f''(x) = \cos(x) -\frac{1}{10}$$
It's easy to see that $f''(x) < 0 \quad \forall \quad x \in [\pi/2, \pi]$, which means that $f'(x)$ is decreasing in that domain.
Now, observe that : 
$$f'\bigg(\frac{\pi}{2}\bigg) = 1 - \frac{\pi}{20} > 0 $$
$$\text{and}$$
$$f'(\pi) = -\frac{\pi}{10} <0$$
By Bolzano's Theorem, (since $f'(x)$ is continuous on $[\pi/2,\pi]$ and $f'(\pi/2)\cdot f'(\pi) < 0$) the function $f'(x)$ has at least one root on the interval $(\pi/2,\pi)$. Taking into account the fact that $f'(x)$ is strictly decreasing on the domain $[\pi/2,\pi]$ though, this means that this root is unique.
Now, it's easy to see that $f(\pi) > 0$. Combining these with the above, it's evident now that the unique critical point lying in $[\pi/2,\pi]$ will be positive and taking into account that $f(0)=f'(0) = 0$ with the fact that $f'(x) > 0$ in the interval $[0,\pi/2]$ as proven above, we can conclude that : 
$$f(x) \geq 0 \quad \forall \quad x \in [0,\pi]$$
and finally, as mentioned, taking into account the property of the function being even, we can conclude that : 
$$f(x) \geq 0 \quad \forall \quad x \in [-\pi,\pi]$$
A numerical analysis approach :
We take the derivative of $f(x)$, which is : 
$$f'(x) = \sin(x) - \frac{x}{10}$$
To find the critical points, we need to solve the equation : 
$$f'(x) = 0 \Rightarrow \sin(x) - \frac{x}{10} = 0$$
but this question does not have a closed form solution. 
Let : 
$$g(x) = \sin(x) - \frac{x}{10}$$
Using Newton's-Raphon's method, we get the equation : 
$$x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)} \Rightarrow x_{n+1} = x_n - \frac{\sin(x_n) - \frac{x}{10}}{\cos(x_n) - \frac{1}{10}}$$
One must be careful though, to pick the correct initial values, which may be tricky. I'll just pick values close to the edges of the interval of study $[-\pi,\pi]$, picking the initial starting points of $x_0 = \pm 2$. Then the method converges successfully and quadratically to the roots of the solution, which are : 
$$x \approx 2.85234$$
$$x \approx - 2.85234$$
Proving that these are the unique roots in the intervals $[-\pi,0)$ and $(0,\pi]$ is a calculus task which has been elaborated in the calculus solution above as well.
Finally, one can pick that $f'(0)=0$ which means that $x=0$ is a solution as well. This can also be done by picking an initial point closer to the mid-point to $0$ of the intervals $[-\pi,0]$ and $[0,\pi]$, like for example $x_0=1$. The method will then converge to $0$, but it will take more time (why ?).
Finally, out of these, we have that $f'(x)$ has $3$ roots all of which are greater than $0$, which means that the critical points are greater than $0$ in the interval given. Thus :
$$f(x) \geq 0 \quad \forall \quad x \in [-\pi,\pi]$$
You can check the calculations of the method here!
More information about the method can be found all over the internet, regarding proofs, expalanation, examples and tricky cases (how to pick correct points, when it coverges and when not) etc.
A: HINT
Since
$$1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24}$$
Thus
$$f(x)=1-\cos x -\frac{x^2}{20}\geq 1-1+\frac{x^2}{2}-\frac{x^4}{24}-\frac{x^2}{20}=\frac{9x^2}{20}-\frac{x^4}{24}\geq 0 \quad x\in[-\pi,\pi]$$
Indeed
$$\frac{9x^2}{20}-\frac{x^4}{24}=x^2\left(\frac{9}{20}-\frac{x^2}{24}\right)\geq0 \iff x^2 \leq \frac{216}{20}=\frac{54}{5} \iff \\x\in \left[-\sqrt{\frac{54}{5}},\sqrt{\frac{54}{5}}\right]\implies x\in\left[-\pi,\pi\right]$$
A: $$1-\cos x=2\sin^2
\frac x2.$$
We need
$$\sin^2 \frac x2\ge\frac{x^2}{10}$$
for $|x|\le\pi$, equivalently
$$\sin^2y\ge\frac{2y^2}5$$
for $|y|\le\frac\pi2$. Again this is equivalent to
$$\sin y\ge\frac{y}{\sqrt{5/2}}\tag1$$
for $0\le y\le\frac\pi2$.
But it is well-known that
$$\sin y\ge\frac{y}{\pi/2}\tag2$$
for $0\le y\le\frac\pi2$ (basically convexity). Then (1) follow from (2)
provided that $\pi/2\le\sqrt{5/2}$, equivalently $\pi^2\le10$.
You can check this on your pocket calculator.
