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V2-x goes next to =x

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

it does
but the end point is not giving

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

so is infinitY/

ok

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

and x is next to V2-x

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

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

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

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

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

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

i don't got a graphing calculator

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

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

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

i am using schoool computer , cannot dowload the.

ok, is this a calculus class?

pre

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

did you dowload the graph program?

yeah i used it to graph your function:)

what did you get?

the domain, is x<_2 , right?

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

thats because of the x in front

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

what are the point of the maximun

cordinate?

so it be (1.37,?

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

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

yes something like that

i need the exact point.

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

wow
y = 1.088