Question

Absolute Max of...

Answer 1

could you find the critical points first?
by equating the 1st derivative to 0 ?

Answer 2

what do you get as first derivative f'(x) =... ?

Answer 3

@hartnn
\[f \prime (x)= \frac{ x^2+4x-1 }{ (x+2)^2 }\]
Critical points were: \[-2 \pm \sqrt{5}\]

Answer 4

:O
i am seeing f(x) =7x-14 sin x!!

Answer 5

haha oops wait a minute...

Answer 6

the first derivative is 7-14cos(x)

Answer 7

as for the critical points im stuck :/

Answer 8

critical points,f'(x)=0
so, for 7-14cos x =0
cos x =... ?

Answer 9

Im getting:\[x=\frac{ \pi }{ 3 }\pm2\pi n, \frac{5 \pi }{ 3 }\pm2\pi n\]

Answer 10

but i asked cos x =... :P

Answer 11

anyways, cos x =1/2
for which angle is cos x =1/2 in your given closed interval ?

Answer 12

the closed interval given is [-pi/4, pi/2]

Answer 13

there is only one angle in that range for which cos x =1/2
you still there?

Answer 14

yes still here sorry! bad internet connection Cos x would = 1/2 @ Pi/3

Answer 15

correct.
so you will now test your function at 3 points,
x=pi/3
x=-pi/4
x=pi/2

for whichever angle , you get f(x) as maximum, is your maximum point :)
can u test ?

Answer 16

yes one second :) let me work it out

Answer 17

\[x=\pi/3; f(x)= \frac{ 7\pi }{ 3 }-7\sqrt{3}\]
\[x=-\pi/4; f(x)= -\frac{ 7\pi }{ 4 }+7\sqrt{2}\]
\[x=\pi/2; f(x)= \frac{ 7\pi }{ 2 }-14\]

Answer 18

so f(x) is max at which x ?

Answer 19

i want to say at x=-pi/4 ?
:|

Answer 20

that is correct :)

Answer 21

Ahhhhh :) yay I actually just adjusted my window on my calculator and it definitely helped idk why i didnt do that in the first place!

Answer 22

cool, and what about the other question, did u get that too ?

Answer 23

not yet unfortunately! :/

Answer 24

what was f(x) ? and u need to find maximim only ?

Answer 25

well this was the original question...

Answer 26

C is the one i am having trouble with

Answer 27

a,b, d, & e i got correct -_-

Answer 28

if f"(x) >0
then its concave up, whats your f"(x) =... ?

Answer 29

\[f \prime (x)=\frac{ 1 }{ 7 }(8x ^{1/7}-x ^{-6/7})\]

Answer 30

\[f \prime \prime (x)=\frac{ 2(4x+3) }{ 49x ^{13/7} }\]

Answer 31

so, that should be > 0
so, x > -3/4
(-3/4, infinity) is not a choice.
i think the 3rd choice has a typing mistake, it didn't show the negative sign...

Answer 32

yes i dont understand. unless my second derivative is incorrect?

Answer 33

your 2nd derivative is correct :)

Answer 34

well what the heck haha. On my calculator at x=0 the graph vertically overlaps the y-axis which is weird

Answer 35

graph of f(x) ? yes, your calculator is weird :P

Answer 36

hmph. so what your saying is the correct answer would be the third choice?

Answer 37

what i was saying is that the correct answer (-3/4, infinity) is not in the choices.
so, it seems, that there was typing error when choices were given, maybe they really meant C to be (-3/4,infinity.)

Answer 38

oh, well. but the graph here shows that its concave up from 0 to infinity!
http://www.wolframalpha.com/input/?i=x%5E%288%2F7%29-x%5E%281%2F7%29

Answer 39

i see it! my calculator looks nothing like that :/ aggh but you were right that was the correct answer!

Answer 40

oh! i think we missed the point that f(x) is only defined for x>= 0 !
so, the interval -3/4 to 0 will not be taken into consideration! thats why its 0 to infinity :)

Answer 41

omg! yes! Thats right ! I get it :O

Answer 42

:)

Answer 43

Thank you so much! I appreciate your help very much!

Answer 44

you're most welcome ^_^