Question

Find the local and absolute extreme values of the function on the given interval.
f(x)=x/(2x^2+5), [0,2]

Answer 1

you got your derivative? what's f '(x) =

Answer 2

I need help with the quotient rule was this the one with the special thing again?

Answer 3

special thing?

Answer 4

hi dee hi hi dee lo thing?

Answer 5

if \(\huge f(x)=\frac{u}{v} \) then \(\huge f'(x)=\frac{v\cdot u'-u \cdot v'}{v^2} \)

Answer 6

oh yeah... "lo d hi minus hi d lo all over lolo"

Answer 7

so let u=x, v=2x^2+5...
so u' = 1, v' = 4x

now plug it into that formula for the quotient rule...:)

Answer 8

(2x^2)(1)-(x)(4x)/(2x^2+5)^2

Answer 9

Right?

Answer 10

\[\huge f'(x)=\frac{v\cdot u' -u \cdot v' }{v^2} \]
\[\huge f'(x)=\frac{(2x^2+5)\cdot (1) -(x) \cdot (4x) }{(2x^2+5)^2} \]

Answer 11

Yeah thats what i got!

Answer 12

oh... i thought you wrote something else....

Answer 13

the answer you wrote didn't have (2x^2+5) in the numerator.

Answer 14

whoops, that is what i wrote down though

Answer 15

where do i go from there?

Answer 16

simplify that numerator... and set it equal to zero and solve...

Answer 17

\[\huge f'(x)=0\rightarrow\frac{5-2x^2 }{(2x^2+5)^2}=0\]
\[\huge 5-2x^2 =0\]

Answer 18

Thats not what i got

Answer 19

did i mess up on the simplification of the numerator?

Answer 20

oh ok i just did it wrong

Answer 21

so solving that will give you two values, but only 1 of them is in the interval [0, 2]

Answer 22

how did you get two values, i only got one i got x=1.5811

Answer 23

is it 1.6 and -1.6?

Answer 24

\[\huge 5-2x^2 =0\]
\[\huge 5=2x^2\]
\[\huge \frac{5}{2}=x^2\]

Answer 25

\[\huge \frac{5}{2}=x^2\]
\[\huge \sqrt{\frac{5}{2}}=x^2\]

Answer 26

sorry... i got the aw snap message...

Answer 27

instead of putting your answer as 1.6, i think it's better to keep it in exact form so at least you don't lose any accuracy.

Answer 28

still there?

Answer 29

yeah sorry

Answer 30

but what do I do?
is that right? 1.581139?

Answer 31

yes... is it ok to put your answers in approximate decimal form instead of the exact form?

Answer 32

I guess. I haven't really asked

Answer 33

but is that the final answer, we dont have to do anything else?

Answer 34

no... remember the question asks for local and global extrema...

Answer 35

yeah

Answer 36

all we did is find where f'(x) = 0....

Answer 37

oh ok :) just kidding

Answer 38

if you don't mind, i'll use the exact form of our answer... \(\large \sqrt{\frac{5}{2}}=\frac{\sqrt{10}}{2} \)

Answer 39

ok

Answer 40

ok... so we need to calculate all of these
\[\huge f(0)= \]
\[\huge f(2)= \]
\[\huge f(\frac{\sqrt{10}}{2})= \]

Answer 41

the first two is just the function evaluated at the endpoints of the interval you're looking at...

Answer 42

uh huh

Answer 43

f(0)= 0

Answer 44

after you have those y-values, you can determine which are local/global max/mins.

Answer 45

f(2)= 2/13

Answer 46

correct..., correct.

Answer 47

umm and im not sure how to figure out the last one

Answer 48

\[(\sqrt{10}/2 )^{2}\] what does that equal

Answer 49

\[\huge f(x)=\frac{x}{2x^2+5}\rightarrow \frac{\frac{\sqrt{10}}{2}}{2(\frac{\sqrt{10}}{2})^2+5} \]
\[\huge \frac{\frac{\sqrt{10}}{2}}{2(\frac{\sqrt{10}}{2})^2+5} =\frac{\frac{\sqrt{10}}{2}}{2(\frac{10}{4})+5}=\frac{\frac{\sqrt{10}}{2}}{5+5} \]

Answer 50

oh ok!

Answer 51

\[\huge \frac{\frac{\sqrt{10}}{2}}{5+5}\rightarrow \frac{\frac{\sqrt{10}}{2}}{10}\rightarrow \frac{\sqrt{10}}{20} \]

Answer 52

so what do you have as far as local/global max/min in your interval [0,2] ?

Answer 53

umm im not sure.. whats the difference in the two

Answer 54

a global extrema is the highest or lowest within the interval...
a local extrema is only the highest or lowest within a "neighborhood"...

Answer 55

so in other words, all global extremas ARE local extremas... but not the other way around...

Answer 56

oh so would the local minimum would be 0?

Answer 57

yes.. x=0 is a local minimum... BUT notice it is the lowest y-value in that interval so it is your global minimum...

Answer 58

keep going... analyze the other two...

Answer 59

what's the highest y-value?

Answer 60

so the global max is \[\sqrt{10}\div20\]

Answer 61

yes...

Answer 62

what about at x=2?

Answer 63

which would that be?

Answer 64

i'm asking you that...

Answer 65

im not sure that

Answer 66

because we still have to find the local min and max and there is only on number left

Answer 67

here's an easy "cheat" way of doing it... look at the graph of the points we have...
if we were to connect them with a smooth curve what can you tell me?