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## ksandoval 3 years ago use analytic methods to find the extreme values of f(x)= (1/x) + lnx on the interval 0.5 ≤ x ≤ 4 and where they occur

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1. ksandoval

and i know that the derivative is f'(x) = -1/x^2 + 1/x but i dont know where to go from there... lol

2. baldymcgee6

Where do extreme values occur?

3. ksandoval

i dont know thats what i need help finding.. lol. i just dont know how to find them.

4. baldymcgee6

okay, so in theory, extreme values will occur where the derivative of the function is equal to zero (i.e. a horizontal slope where there is maxima or minima), and they also occur where the derivative is undefined.

5. ksandoval

okay sooo i set the derivative equal to zero...

6. baldymcgee6

and...?

7. ksandoval

well i mean plugging in 1 for x would give you zero.

8. baldymcgee6

okay, so 1 is one of our critical points

9. ksandoval

now what do we do?

10. baldymcgee6

Where is f'(x) = -1/x^2 + 1/x undefined?

11. ksandoval

when x = 0?

12. baldymcgee6

exactly, but if you would notice in the question it gave us restrictions of 0.5 ≤ x ≤ 4, so we dont take 0 into account... So our critical point is 1

13. baldymcgee6
14. ksandoval

oooh ok. so for the answer it says: max value is 1/4 + ln4 at x = 4 min value is 1 at x = 1 local max at (1/2, 2 - ln2) how did they get the max value and local max?

15. baldymcgee6

Well for the max value, they just put the biggest number they could, 4, into the function, they chose 4 because of the restrictions 0.5 ≤ x ≤ 4... 4 is the biggest number, i.e. giving the biggest value.

16. baldymcgee6

there is no local max... so i'm not sure where they got that.. might want to ask your teacher.

17. ksandoval

oh.. ok. and how do they know that 1 is the min value? this is confusing for me ):

18. baldymcgee6

We already got that the minimum is at x = 1

19. ksandoval

yea but how do you know its the minimum? :\

20. baldymcgee6

We know it is the minimum because at that point, the derivative = 0, this is the lowest point on the curve.

21. ksandoval

thank you

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