Question

Find the absolute maximum and minimum for y=x^2+(2/x) on 0.3 < or equal to x < or equal to 4

Answer 1

\[y=x^{2}+\frac{ 2 }{ x }\]

Answer 2

take the first derivative and solve for zero is step one

Answer 3

I am not sure where to start so I need help

Answer 4

That will give you x=0 and x=\[\sqrt[3]{2}\]

Answer 5

To get a maximum the first derivative must go from increasing to decreasing. When the first derivative reaches zero, it sometimes changes from inc. to dec. So I would set up a numberline with a value below zero, a value between 0 and \[\sqrt[3]{2}\] and a value greater than \[\sqrt[3]{2}\

Answer 6

ok

Answer 7

You with me so far?

Answer 8

yes

Answer 9

OK, now set up your numberline and tell me what you see

Answer 10

I see one above and one below

Answer 11

so the value below zero is a pos. number?

Answer 12

yes

Answer 13

that means that the function is inc. from neg. inf. to zero.

Answer 14

right

Answer 15

The value between zero and the cubed root of 3 is a neg.

Answer 16

ok

Answer 17

that means that the function is dec. To have a max. a function must go from inc. to dec. therefore, your function has a max. at x=0.

Answer 18

We aren't done yet, what is the value after the cubed root of 3?

Answer 19

so then absolute ax is 0

Answer 20

Im pretty sure that is the case on this interval.

Answer 21

ok

Answer 22

is the value after the cubed rood of 3 a pos. number

Answer 23

yes

Answer 24

then that is your abs. min. because the function is going from dec. to inc. there

Answer 25

Before you call it quits, check the values of your endpoints.

Answer 26

can you just tell me what you calculated for max and min just to match my number s with yours to ake sure I got the right thing

Answer 27

my first derivative was 2x-2/x^2

Answer 28

yup got that

Answer 29

what value did you choose for below zero? I'm getting neg. 4 for my value below zero. I could be wrong since i don't have anything to check my work.

Answer 30

not I got -4 too

Answer 31

that means the function is dec. there

Answer 32

right

Answer 33

i think i may have made a math error when I solved for my zeros. my x=0 is still good but instead of the cubed root of 3, it should be just 1. I'm sorry. But that shouldn't throw us off too much. Now im going to plug in 1/2 for my second value

Answer 34

ok I may get a litle confused but I will try and correct what you just said

Answer 35

My second value gives me 8, a positive number. Your function has a min. at x=0

Answer 36

ok I folowing so far

Answer 37

for my third value, im going to plug in 2 because it is easy to do the math. and that will give me three halves

Answer 38

ok

Answer 39

so we still have an inc. function. Now we plug in the end points and our min. into the original funciton. The lowest number will be our abs. Min. and our highest number will be our abs. Max.

Answer 40

I could be wrong, Im only a BC calculus student. If I had my trusty TI-89 I could be absolutely sure. I would check with someone who knows what they are doing before I turn this in.

Answer 41

ok I thnks there is an onine graphing calculator that you can use

Answer 42

Im getting opposite numbers so I don't think I didi it right

Answer 43

Yes but it doesn't like me and it won't do what i tell it to. My TI-89 is the only calculator i really trust because i know exactly how to use it

Answer 44

well do you know anybody that could help us

Answer 45

No you probably did it right, I lack a calculator to back me. Now where are you getting opposite numbers. what intervals

Answer 46

I'm assuming you are a calculus student. Am I right?

Answer 47

Try to get a hold of one of your classmates to confirm what you have done. I think this problem is almost solved. If I had all of my tools, We'd have this thing done already. I really should have brought my materials with me over spring break. Anyway, If you have any questions pertaining to calculus, Tutorial math is an excellent source. Im in second semester calculus and that site is the reason I have a B in there. Good luck with your math assignment.