Question

need help on the attachment

Answer 1

what attachment?

Answer 2

the one that he prolly cant attach quickly enough cause of OS being on a monthly cycle lol

Answer 3

haha very funny amistre64

Answer 4

:)

so what do you end up getting for your 1st and 2nd derivatives?

Answer 5

i see its a product rule and a power rule to play with

Answer 6

\[x(2-x)^{3/4}\]

Answer 7

product rule and chain rule on the part in parentheses:
\[f'(x)=(2-x)^{3/5}+x(3/5)(2-x)^{-2/5}(-1)\]\[=(2-x)^{-2/5}[(2-x)-3x/5]=(2-x)^{-2/5}(2-8x/5)=0\]

Answer 8

no its (2-x)^(3/5)

Answer 9

you can get the critical numbers from this I presume

Answer 10

\[x'(2-x)^{3/5}+x(2-x)'^{3/5}\]
\[(2-x)^{3/5}+x\frac{3}{5}(2-x)^{-2/5}\]
\[(2-x)^{3/5}+\frac{3x}{5(2-x)^{2/5}}\]
\[5(2-x)^{2/5}(2-x)^{3/5}=5(2-x)\]
\[\frac{3x(5(2-x))}{5(2-x)^{2/5}}\]
\[\frac{3x(2-x)}{(2-x)^{2/5}}\]
\[3x(2-x)^{3/5}=0;x=0,2\]
those look almost critical

Answer 11

i missed it someplcae if wolframs right

Answer 12

you forgot chain rule on the right and lost a negative sign amistre

Answer 13

gpt it; i produced instead of sumate

Answer 14

and the negative too lol

Answer 15

so is the answer suppose to be -2, also amistre64 I don't get your work on step 4

Answer 16

step 4 was just so that I could rewrite the numerator with a common denominator

Answer 17

???

Answer 18

lets start at step 3

Answer 19

since i messed it up afterwards anyhoos :)

Answer 20

ok

Answer 21

\[(2-x)^{3/5}-\frac{3x}{5(2-x)^{2/5}}\]
\[(2-x)^{3/5}*\frac{5(2-x)^{2/5}}{5(2-x)^{2/5}}-\frac{3x}{5(2-x)^{2/5}}\]
\[\frac{5(2-x)}{5(2-x)^{2/5}}-\frac{3x}{5(2-x)^{2/5}}\]
\[\frac{5(2-x)-3x}{5(2-x)^{2/5}}\]
\[\frac{10-5x-3x}{5(2-x)^{2/5}}\]
\[\frac{10-8x}{5(2-x)^{2/5}}=0;\ when\ x=\frac{\cancel{10}5}{\cancel{8}4}\]

Answer 22

and it is undefined at x=2

Answer 23

so these are important parts to check out

Answer 24

im guessing it just wants the 1st derivative criticals

Answer 25

yes I believe so

Answer 26

so try x=2,5/4 and hopefullt thats it :)

Answer 27

where did 5/4 come from

Answer 28

its what the 1st derivative zeros out at

Answer 29

10-8x = 0 when x=5/4

Answer 30

oh, you set the numerator equal to zero

Answer 31

ok so you always set the numerator and denominator equal to zero when solving for the critical points

Answer 32

yes, criticals are when you zero out the top, and where you have undefineds at, or zeros in the bottom

Answer 33

they are points of interest

Answer 34

ok thanks so much

Answer 35

Also one thing how did you get rid of 3/5 on (2-x)?