Two points on graph are equally the largest. Are both global maximums? So I've been given the equation $$f(x)= x^3-4x^2+4x$$ for 0< x< 4
The derivative i've calculated is $$3x^2-8x+4$$
setting it equal to zero gets me the critical points $$\frac23 ,2$$
I've been taught to plug these two back into the equation, along with the interval points.
So f(2/3)= 32/27
f(2)= 0 
f(0) = 0
f(4) = 16. 
So 4 is the maximum but is 0 or 4 the minimum?
Slightly different question on the same subject. What do I do if the interval I'm given is x>0? My range is technically (0,inf). The first can be input. The latter cannot. What would you do? 
 A: First of all, is the question really strict inequality ($ 0 < x < 4$) or is it non-strict inequality ($0 \leq x \leq 4$)? The way you've stated the problem, you don't actually check the points where $x = 0$ and $x = 4$ because they're not actually in the domain.
When we talk about the maximum and minimum of functions on real numbers, we talk about the values achieved in the range (the $y$ values), not the values from the domain (the $x$ values). So, assuming the domain is actually $0 \leq x \leq 4$, the maximum of your function on the given domain is 16 and the minimum is 0.
If the question asks where is the maximum, then you can say you get the maximum when $x = 4$. If it asks where the minimum is, you can say the minimum is at $x = 0$ and $x = 2$.
If the interval ends at infinity, then you don't have to check that bound because there is no bound to check. (Even though we sometimes say "range" to mean "interval", try not to confuse the two words. In the context of the math you're doing, you need to know that the domain is where the function comes from and the range is where the function ends up.)
