anonymous
  • anonymous
Consider the following.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
8
anonymous
  • anonymous
g(x)=x\[\sqrt{2-x}\] Use a graphing utility to approximate the relative minimum or relative maximum value of the function
anonymous
  • anonymous
V2-x goes next to =x

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More answers

amistre64
  • amistre64
square roots dont have a relative min or max..... unless its their endpoints
anonymous
  • anonymous
it does but the end point is not giving
amistre64
  • amistre64
youre going to have to rewrite the question to make sense of it for us :)
anonymous
  • anonymous
so is infinitY/
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[g(x)=\sqrt{2-x}\]
anonymous
  • anonymous
and x is next to V2-x
anonymous
  • anonymous
so is g(x)=xV(2-x)
dumbcow
  • dumbcow
domain is x<=2 relative maximum at x=4/3 i used calculus and set derivative equal to 0
anonymous
  • anonymous
relative maximum at x=4/3, why not a cordinate?
dumbcow
  • dumbcow
to get y coordinate substitute 4/3 for x y = 4/3 * sqrt(2-4/3)
anonymous
  • anonymous
so you get, g(4/3)=4/3V(2-3/4)
dumbcow
  • 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
  • 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
  • dumbcow
no they dont but g(x) does because it has the x out front of the sqrt
anonymous
  • anonymous
i don't got a graphing calculator
anonymous
  • anonymous
and i search in the internet,but din't find any that could graph square root functions
dumbcow
  • dumbcow
try this http://www.graphcalc.com/download.shtml
anonymous
  • anonymous
i got a similar problem, with the answe,r wanna me to show you?
anonymous
  • anonymous
i am using schoool computer , cannot dowload the.
dumbcow
  • dumbcow
ok, is this a calculus class?
anonymous
  • anonymous
pre
dumbcow
  • dumbcow
ok for last problem you def need the graph or plot the points from 0 to 2 to see the maximum
anonymous
  • anonymous
did you dowload the graph program?
dumbcow
  • dumbcow
yeah i used it to graph your function:)
anonymous
  • anonymous
what did you get?
anonymous
  • anonymous
the domain, is x<_2 , right?
dumbcow
  • dumbcow
yes here is the graph
1 Attachment
anonymous
  • anonymous
oh, it looks different from what i thought it would be .
dumbcow
  • dumbcow
thats because of the x in front
anonymous
  • anonymous
nice., so what does the x do? slope>?
anonymous
  • anonymous
?
anonymous
  • anonymous
what are the point of the maximun
dumbcow
  • 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
  • anonymous
cordinate?
dumbcow
  • 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
  • anonymous
so it be (1.37,?
anonymous
  • anonymous
use trace on the graph, to find the highest point.
anonymous
  • anonymous
use trace on the graph, to find the highest point.
dumbcow
  • dumbcow
yes something like that
anonymous
  • anonymous
i need the exact point.
dumbcow
  • 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
  • anonymous
don't got a caclculator iwth me, bye, i will solve it when ig et home.
dumbcow
  • dumbcow
wow y = 1.088

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