Grade upto college level Integral Calculus
f(x)=1+2sin x+3cos ²x[0,2p÷3]
FIND MAX VALUE OF FUNCTION
2 Answers
Sunil Raikwar
Approved Tutor Answer12 Years agoLet y = 1+2sinx+3cos2x
Differentiate with respect to x
y' = 2cosx-6cosxsinx or 2cosx-3sin2x
put y' = 0
cosx(1-3sinx) = 0
cosx = 0, sinx = 1/3
again Diffentiate with respect to x
y'' = -2sinx-6cos2x
y'' greater than zero for cosx=0 & less than zero for sinx=1/3
hence y give maximum value at sinx=1/3
maximum value of y= 1+ 2/3+ 3*2/9=5/3
Sunil Raikwar
Let y = 1+2sinx+3cos2x
Differentiate with respect to x
y' = 2cosx-6cosxsinx or 2cosx-3sin2x
put y' = 0
cosx(1-3sinx) = 0
cosx = 0, sinx = 1/3
again Diffentiate with respect to x
y'' = -2sinx-6cos2x
y'' greater than zero for cosx=0 & less than zero (negative) for sinx=1/3
hence y give maximum value at sinx=1/3
maximum value of y= 1+ 2/3+ 3*2/9=5/3
Thanks & Regards
Sunil Raikwar
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