If f(x)=1/x and I want to find the absolute min/max in the range [-1, 1], is the answer -infinity and +infinity? I concluded this by drawing the graph, but is there a mathematical method to find the answer on such case? Or must the function be continius at the range to find absolute min/max?

Thanks a lot!

try limits!
(you see, the function is not continuus)

he means [-1,0) and (0,1] :cool:

To find the local maxima and minima put f'(x) = 0.

In this case f'(x) = -1/x²

This is zero when x = +/- infinity. That gives the minimum values for the function (where f(x)=0).

f'(0) is undefined at x=0, so we need to check that point separately. f(x) at x=0 is also undefined, as it happens.

We then consider the limit as x approaches zero from the left or right. In that case, f(x) goes to -infinity or +infinity.

In fact we don't have to look at f' to get the answer...

if we look at y=1/x, we have to conclude that the bigger the value of x is, the smaller the value of y.
And....the smaller the value of x, the bigger the value of y.
We all know there ain't a 'smallest' value for x, so there can't be a maximum for y: infinity.

+ the fact that when we divide something with 0, there can't be an answer, hence x=0 doesn't give us a value for y.

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hmmm, thinking of 1/x, the point of bowing f''=0 can't be solved...
which ain't logical since the graphic does 'bow':

f''=2x/x^4
2x/x^4=0

this means x ain't 0, to divide something bij 0 gives an impossible answer.
but is also says x=0, when you divide and the answer is 0; 2x MUST be 0.
Kinda impossible and strange huh...

is there any other graphic which bows, but gives an f'' which is impossible to solve. (or did I make a miscalculation somewhere?)

well, 2x/x^4 is reall 2/x^3, which makes it unsolvabe if you want to get 0.

That's a better explanation of what I tried to say:D
thx:cool: