# max and min point

• November 7th 2012, 11:15 AM
Petrus
max and min point
solve max and min https://webwork.math.su.se/webwork2_...e864664931.png range https://webwork.math.su.se/webwork2_...40ef46cb01.png
my progress derivate it and get 1/x+1-2x and find the roots x=1 and x=1/2 what is my next step?
need a tips what i shall do next
• November 7th 2012, 11:20 AM
fkf
Re: max and min point
Quote:

Originally Posted by Petrus
solve max and min https://webwork.math.su.se/webwork2_...e864664931.png range https://webwork.math.su.se/webwork2_...40ef46cb01.png
my progress derivate it and get 1/x+1-2x and find the roots x=1 and x=1/2 what is my next step?
need a tips what i shall do next

We have
f(x) = ln(x)+x-x^2 => f'(x) = 1/x+1-2x

Searching for max/min by finding for which x we have f'(x) = 0. Hence
1/x+1-2x = 0 => x = 1 and x = -1/2

But we're looking in the range [1/4,2] and x = -1/2 is not in that range. x = 1 are in the range, therefore x = 1 is a valid solution to the problem. Now you can check if this is a max or min by using the second derivative or by studying if the first derivative is positive or negative around x = 1.
• November 7th 2012, 11:30 AM
Petrus
Re: max and min point
Quote:

Originally Posted by fkf
We have
f(x) = ln(x)+x-x^2 => f'(x) = 1/x+1-2x

Searching for max/min by finding for which x we have f'(x) = 0. Hence
1/x+1-2x = 0 => x = 1 and x = -1/2

But we're looking in the range [1/4,2] and x = -1/2 is not in that range. x = 1 are in the range, therefore x = 1 is a valid solution to the problem. Now you can check if this is a max or min by using the second derivative or by studying if the first derivative is positive or negative around x = 1.

i enter in the computer empty on min point and i get inorrect x=1 is max point and its correct but how do i solve min =S?
• November 7th 2012, 11:35 AM
fkf
Re: max and min point
Quote:

Originally Posted by Petrus
i enter in the computer empty on min point and i get inorrect x=1 is max point and its correct but how do i solve min =S?

You have the value for f(1). But you also have the range [1/4,2]. Which means that the values of the function lies in this interval. Therefore you also have the values 1/4 and 2. And also the f(1).
• November 7th 2012, 11:37 AM
MarkFL
Re: max and min point
Evaluate the function at the endpoints of the given interval and at the critical value within the interval. The largest of these is your absolute maximum and the smallest is your absolute minimum.
• November 7th 2012, 11:39 AM
Petrus
Re: max and min point
Ty mark!:) Tack fkf :)
• November 7th 2012, 11:40 AM
fkf
Re: max and min point
Quote:

Originally Posted by MarkFL2
Evaluate the function at the endpoints of the given interval and at the critical value within the interval. The largest of these is your absolute maximum and the smallest is your absolute minimum.

Since we have the range, we don't need to calculate the values for the endpoints, since the range is giving us the values for some x.
• November 7th 2012, 11:41 AM
fkf
Re: max and min point
Quote:

Originally Posted by Petrus
Ty mark!:) Tack fkf :)

:) I did edit my post, but maybe you already got the point.
• November 7th 2012, 11:43 AM
Petrus
Re: max and min point
Ty :) you prob also read math in Stockholm university?
• November 7th 2012, 11:45 AM
fkf
Re: max and min point
Quote:

Originally Posted by Petrus
Ty :) you prob also read math in Stockholm university?

Nope, been studying geology there (distance course). I just started to study civil engineering in applied physics (or maybe it's called engineering physics).
• November 7th 2012, 11:57 AM
MarkFL
Re: max and min point
Quote:

Originally Posted by fkf
Since we have the range, we don't need to calculate the values for the endpoints, since the range is giving us the values for some x.

I interpreted the information given that we have $\frac{1}{4}\le x\le2$. I assumed Petrus meant domain instead of range.