Consider the following.

- anonymous

Consider the following.

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- anonymous

8

- anonymous

g(x)=x\[\sqrt{2-x}\]
Use a graphing utility to approximate the relative minimum or relative maximum value of the function

- anonymous

V2-x goes next to =x

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## More answers

- amistre64

square roots dont have a relative min or max..... unless its their endpoints

- anonymous

it does
but the end point is not giving

- amistre64

youre going to have to rewrite the question to make sense of it for us :)

- anonymous

so is infinitY/

- anonymous

ok

- anonymous

\[g(x)=\sqrt{2-x}\]

- anonymous

and x is next to V2-x

- anonymous

so is g(x)=xV(2-x)

- dumbcow

domain is x<=2
relative maximum at x=4/3
i used calculus and set derivative equal to 0

- anonymous

relative maximum at x=4/3, why not a cordinate?

- dumbcow

to get y coordinate substitute 4/3 for x
y = 4/3 * sqrt(2-4/3)

- anonymous

so you get, g(4/3)=4/3V(2-3/4)

- dumbcow

you need to find a graphing calculator or grAPHING COMPUTER software to see it visually
which i think is what the question is asking

- anonymous

yeah, the is what it asking
but like somebody above said, square root fuction don't have a maximun or minimum. Then how you find it?

- dumbcow

no they dont but g(x) does because it has the x out front of the sqrt

- anonymous

i don't got a graphing calculator

- anonymous

and i search in the internet,but din't find any that could graph square root functions

- dumbcow

try this
http://www.graphcalc.com/download.shtml

- anonymous

i got a similar problem, with the answe,r wanna me to show you?

- anonymous

i am using schoool computer , cannot dowload the.

- dumbcow

ok, is this a calculus class?

- anonymous

pre

- dumbcow

ok for last problem you def need the graph or plot the points from 0 to 2 to see the maximum

- anonymous

did you dowload the graph program?

- dumbcow

yeah i used it to graph your function:)

- anonymous

what did you get?

- anonymous

the domain, is x<_2 , right?

- dumbcow

yes
here is the graph

##### 1 Attachment

- anonymous

oh, it looks different from what i thought it would be .

- dumbcow

thats because of the x in front

- anonymous

nice., so what does the x do? slope>?

- anonymous

?

- anonymous

what are the point of the maximun

- dumbcow

kinda, it changes the direction of the graph
when sqrt(2-x) might be decreasing as x increases
when you multiply by x, it increases to a point
the max is the highest point on graph, x=4/3

- anonymous

cordinate?

- dumbcow

i told you earlier, plug in 4/3 into function
look at cursor position on right side of picture
i left it at approx where the max is

- anonymous

so it be (1.37,?

- anonymous

use trace on the graph, to find the highest point.

- anonymous

use trace on the graph, to find the highest point.

- dumbcow

yes something like that

- anonymous

i need the exact point.

- dumbcow

from a graph you will only get an approximate but thats all you need right now
the exact answer i already told you
max: (4/3, 4/3*sqrt(2-(4/3)))

- anonymous

don't got a caclculator iwth me, bye, i will solve it when ig et home.

- dumbcow

wow
y = 1.088

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