Minimum of $x^2 + 3x - 1$ on $[0,1]$ and $[-2,2]$ Consider the problem of finding the absolute minimum of the function $f : [0,1] \rightarrow \mathbb{R}$ that satisfies $f(x)=x^2 + 3x - 1$ everywhere.
Suppose we suspect, by graphical methods, that $f$ attains its absolute minimum at the point $0 \in [0,1]$. Thus its absolute minimum is believed to equal $f(0)=-1$.
How would you prove, in a rigorous and logically explicit manner, that $f$ really does achieve its absolute minimum at the point $0$ in its domain? Please, make your premises explicit and identify all the major theorems you use.
Consider also the function $g : [-2,2] \rightarrow \mathbb{R}$ with the same defining equation. Suppose you believe that $g$ attains its absolute minimum at $-3/2$ (which is a local minimum), thus its absolute minimum is $g(-3/2)=-13/4$. The same question applies: how would you confirm your suspicions?
 A: For any $x\in [0,1]$, $f(x)=x^2+3x-1\geq 0+0-1=-1$.
For any $x\in [-2,2]$, $g(x)=x^2+3x-1=(x+\frac 32)^2-\frac{13}{4}\geq 0-\frac{13}{4}=-\frac{13}{4}$.
A: Jasper's answer is for your case for sure much better than mine. This is a much more complicated answer, which is more general and can be applied in more situations.
We know that every continuous function attains its minimum and maximum on a compact set.
Furthermore we know that a polynomial is differentiable, and when a function is differentiabe and takes a minimum in the interior the derivative must be zero at this point. But $2x+3$ has no zero in $[0,1]$. So the minimum must be at $0$ or at $1$. As $f(0)=-1$ and $f(1)=3$, we know that $f$ attains its minimum at $x=0$.
For the second case, you can do what Jasper says or this one which is much more general. When $f$ is convex and $f$ attains a local minimum at $x_0$ it is the global minimum.
A function is called convex if for all $t \in (0,1)$  the following inequality holds:
$$f(t\cdot x_0 + (1-t) x_1)\leq t\cdot f(x_0) + (1-t) f(x_1)$$ 
It sounds pretty complicated doesn't it? It just says that the values of the function are below of the secant of two points.
When $f$ is twice differentiable it is convex iff $f''\geq 0$ everywhere. 
Obviously our function is convex with the remark. 
A function $f$ attains a local minimum at $x_0$ if there exists a neighbourhood with an $\varepsilon >0$ such that for all $x$ with $|x_0-x|<\varepsilon$ the inequality 
$$f(x)\geq f(x_0)$$
holds.
A function $f$ attains a global minimum at $x_1$ if for all $x$ 
$$f(x_1)\leq f(x).$$
I just told all that because I am not sure if the asker knows those definitions in that way. Sorry if someone got bored. 
So now we say that $f$ attains its local minimum at $x_0$ but $x_0$ is not the global minimum, so there must be an $x_2$ such that $f(x_2)< f(x_0)$ (note that $x_2$ doesn't need to be a minimum or something like that). We know that $f$ is convex so we know that 
$$f(t \cdot x_0+ (1-t) \cdot x_2)\leq t \cdot f(x_0) + (1-t)\cdot f(x_2)$$
But as we know that $f(x_2)< f(x_0)$ we have 
$$t\cdot f(x_0) + (1-t)\cdot f(x_2)< t \cdot f(x_2) + (1-t)\cdot f(x_2)=f(x_2)$$
so we have 
$$f(t\cdot x_0 +(1-t) \cdot x_2)< f(x_2)$$
for all $t\in (0,1)$, when we take $t$ really really near $1$  we have 
$$|x_0- t \cdot x_0 +(1-t)\cdot x_2|<\varepsilon$$
But we said that $f$ at this point is smaller than $f(x_0)$, hence $x_0$ can't be a local minimum, when it isn't the global minimum. 
But we said that it is a local minimum and so we proved that every local minimum of a convex function is a global minimum.
A: The derivative of $f$ is $f´(x)=2x+3$.
That means that on $(-\infty,-3/2)$ the function $f$ is decreasing because of $f´(x)<0$ and that on $(-3/2,+\infty)$ the function $f$ is increasing because of $f`(x)>0$. So it is obvious that the minimum is attained at $0$ and is equal to $f(0)=-1$.
Practically the same applies to $g$ because $g´(-3/2)=0$ and if $g`(x_0)=0$ then $x_0$ could be minimum but it doesn´t necessarily needs to be but for this $g$ it is also the case that $g´´(-3/2)=3$ so the second derivative shows that it is really minimum. You only need basics of differential caluculus, call it analysis if you want.
If you want theorems which are used, as you say, then:
1) If $f$ is differentiable on $(a,b)$ and $f`(x)<0$ on $(a,b)$ then $f$ is decreasing on $(a,b)$
2) If $f$ is differentiable on $(a,b)$ and $f`(x)>0$ on $(a,b)$ then $f$ is increasing on $(a,b)$
3) If $f$ is differentiable at the point $x_0$ nad $f`(x_0)=0$ then $x_0$ could be minimum, or maximum, or saddle point. To deduce the character of the point $x_0$ you need second derivative or sometimes even higher derivatives.
A: Critical Points
Suppose $f:[a,b]\rightarrow \mathbb{R}$ is continuous. Call $x\in[a,b]$ is a critical point for $f$ on $[a,b]$ if either


*

*$x$ is an endpoint.

*$x$ is a stationary point: $f'(x)=0$.

*$f$ is not differentiable at $x$.


Proposition: Closed Interval Method
Let $f:[a,b]\rightarrow\mathbb{R}$ be continuous. If $f$ attains it absolute maximum or absolute minimum at some point $x\in[a,b]$, then $x$ is a critical point for $f$ on $[a,b]$.
Proof: Suppose $f$ attains an absolute extremum at $x\in [a,b]$. If $x$ is an endpoint then $x$ is a critical point. Therefore suppose $x\in(a,b)$ and suppose further that $f$ is differentiable at $x$. Then $f'(x)=0$ and so $x$ is a critical point. If $f$ is not differentiable at $x$ then $x$ is still a critical point $\bullet$
Remark
Therefore to find the absolute maximum/ minimum of a continuous function on a closed interval, find the critical points $\{x_1,x_2,\dots,x_n\}$ and evaluate the function at each. Then the absolute maximum/ minimum is the maximum/ minimum of $\{f(x_i):i=1,\dots,n\}$.
