Hmm the derivative is
Now you can see that over [0,3], it's always positive. So f is always increasing.
Meaning, as you said , that the min is at x=0 and the max is at x=3.
Be careful, it's not always the case !! This works under given conditions only !
The best way to check is probably to draw a graph
hmmmm, i get it a bit, but i see why you made a point there.
ummm , i know it seems i ask too much :/
but this chapter confuses me alot,
especially when they throw me off with trig functions and especially exponents.
im currently stuck with this one:
all ive done so far is
but i dont know what to do next
Nope, the derivative is
Note that if x=0, the derivative is not defined. Though the function exists.
Since the derivative would be infinity, this means that there is a vertical asymptote to the curve of f. Sketch it on your calculator
Then you can see that if you consider (0,infinity) (0 not included), . So f is strictly increasing over [0,infinity) (0 is included - because f is defined at x=0), meaning that 0 is a minimum.
The maximum is . so there is no maximum.
For (-infinity,0), you can just note that , because
So 0 is also a minimum value for f over (-infinity,0]
I really hope this helps you grasping the problems... I'm not that good at explaining these things