Minimum of a Function 
Let $I := [0, \frac{\pi}{2}]$ and let $f : I \to \Bbb{R}$ be defined by $f(x) = \sup \{x^2, \cos(x)\}$ for $x \in I$. Show there exists an absolute minimum point $x_0 \in I$ for $f$ on $I$. Show that $x_0$ is a solution to the equation $x^2 = \cos(x)$. 

I am having a little trouble with this problem. From the max/min theorem, we know that such an $x_0$ described above exists; moreover, by the intermediate value theorem it is clear that there exists a $z \in I$ such that $z^2 = \cos(z)$. My strategy is to show that $x_0 = z$. Since $f(x_0)$ is minimum, we know that $\sup \{x_0^2, \cos(x_0) \} \le \max \{z^2, \cos(z)\} = z^2$. Hence $x_0^2 \le z^2$, $\cos(x_0) \le z^2$, and $\cos (x_0) \le \cos(z)$, and therefore $(x_0 - z)(x_0 + z) \le 0$. 
Now, if $x_0 \neq z$ were true, then $x_0 + z > 0$ and so we would need $x_0 - z \le 0$ or $x_0 \le z$. Since $\cos()$ is decreasing on $I$, $\cos (z) \le \cos (x_0)$, which contradicts the fact that $\cos (x_0) \le \cos (z)$. However, I don't yet have access to the concept of a derivative yet to show that $\cos()$ is decreasing. 
I am not sure how to finish this problem without the assumption that $\cos()$ is decreasing on $I$. I could use some hints.
 A: The argument works for the general case of $f(x)=\sup\{g(x),h(x)\}$ where $g,h\colon [a,b]\to \Bbb R$ are continuos, $g$ strictly increasing, $h$ strictly decreasing, $h(a)>g(a)$, and $g(b)>h(b)$:
A minimum point $x_0\in [a,b]$ exists because $[a,b]$ is compact and $f$ is continuous.
If $f(x_0)\ne g(x_0)$, then $f(x)\ne g(x)$ in a neighbourhood of $x_0$. But then $f(x)=h(x)$ in that neighbourhood. As $h$ is strictly decreasing, this is only possible if said neighbourhood does not extend to the right of $x_0$, i.e., $x_0=b$. But then $f(x_0)=g(x_0)$, contradiction. We conclude $f(x_0)=h(x_0)$.
The proof for $f(x_0)=g(x_0)$ is essentially the same.
Note that you do not need derivatives to show that $\cos x$ is strictly decreasing (or that $x^2$ is strictly increasing):
If $0\le x<y\le\pi$ then $\cos y-\cos x=-2\sin\frac{x+y}{2}\sin\frac{y-x}2<0$ because $0<\frac{y-x}{2}<\frac{x+y}2<\pi$.
A: so in a $2$ element set, the supremum is basically the maximum.
IF they are both equal, then it's the value.
So basically, as you can see, $f(x)=\cos(x)$ on the interval $(0,x_0]$ because $\cos(x)\geq x^2$.
And $f(x)=x^2$ on the interval $[x_0,\frac{π}{2})$ because $x^2 \geq \cos(x)$.
Consider $f(x)$ near $x=x_0$. From the left $f(x)$ is decreasing, and from the right $f(x)$ is increasing.
This change in sign of the derivative signals a relative minimum.
And right as $\cos(x)$ and $x^2$ pass each other on the number line, the function switches pieces. This is also where $\cos(x)=x^2$, which is when $x=x_0$. 
