anonymous
  • anonymous
The maximum value of the function
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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hartnn
  • hartnn
can be found by taking the derivative of that funtion and equating it to 0.
anonymous
  • anonymous
y=\[\sqrt{2-x}\]
hartnn
  • hartnn
so did u find y' ?

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anonymous
  • anonymous
y=-4[\sqrt{2-x}\]
hartnn
  • hartnn
could u differentiate y ?
anonymous
  • anonymous
differentiate the function, then put the differentiated equation = 0 you will have either one answer or 2, if you get two answers, the one which has negative value will be the maximum value of the function ;)
anonymous
  • anonymous
is it the y value?
hartnn
  • hartnn
have u learned to differentiate x^n ?
anonymous
  • anonymous
y'= nx^n-1
hartnn
  • hartnn
so couldn't u differentiate y=-4 sqrt(2-x) ??
anonymous
  • anonymous
does the max. value asked here is the x or the y value?
hartnn
  • hartnn
y value
anonymous
  • anonymous
y'= 2/(2-x)^-(1/2)
anonymous
  • anonymous
@dan18 , the correct way is to differentiate and equate to zero , but just by looking at this question once can tell it's maximum value is infinity for x=-infinity....pretty weird question i guess...are u sure u hv typed the question correctly
hartnn
  • hartnn
wouldn't it be y'= 2/((2-x)^(1/2))
anonymous
  • anonymous
I get x=2 @ y', then subts. it to the eq. does the max. value = 0?
anonymous
  • anonymous
@dan18 is the question correct ?
hartnn
  • hartnn
but actual answer is infact Max. value = 0 when x=2 http://www.wolframalpha.com/input/?i=y%3D-4+sqrt%282-x%29
anonymous
  • anonymous
\[y=-4\sqrt{2-x}\]
anonymous
  • anonymous
thak you guys!
anonymous
  • anonymous
now u hv got the correct question, yes the max would be zero

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