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8
g(x)=x\[\sqrt{2-x}\] Use a graphing utility to approximate the relative minimum or relative maximum value of the function
V2-x goes next to =x

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Other answers:

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)
you need to find a graphing calculator or grAPHING COMPUTER software to see it visually which i think is what the question is asking
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?
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?
yes here is the graph
1 Attachment
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
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
cordinate?
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
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.
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)))
don't got a caclculator iwth me, bye, i will solve it when ig et home.
wow y = 1.088

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