dan18
The maximum value of the function
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hartnn
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can be found by taking the derivative of that funtion and equating it to 0.
dan18
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y=\[\sqrt{2-x}\]
hartnn
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so did u find y' ?
dan18
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y=-4[\sqrt{2-x}\]
hartnn
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could u differentiate y ?
miteshchvm
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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 ;)
dan18
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is it the y value?
hartnn
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have u learned to differentiate x^n ?
dan18
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y'= nx^n-1
hartnn
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so couldn't u differentiate y=-4 sqrt(2-x) ??
dan18
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does the max. value asked here is the x or the y value?
hartnn
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y value
dan18
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y'= 2/(2-x)^-(1/2)
akash_809
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@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
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wouldn't it be y'= 2/((2-x)^(1/2))
dan18
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I get x=2 @ y', then subts. it to the eq. does the max. value = 0?
akash_809
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@dan18 is the question correct ?
dan18
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\[y=-4\sqrt{2-x}\]
dan18
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thak you guys!
akash_809
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now u hv got the correct question, yes the max would be zero