f(x)= cos x/2+sin x ;

f(x)= (2+sin x)(-sin x)-cos x cos x/`(2+sin x)^(2)`

-- using the tool of calculus, it can be shown that function f(x) given, the zeroes of f'(x) give the location of any maximum and/or minimum values. find the location of these values in the inteval [0,2 ], using trig identities as needed to solve f'(x) = 0

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`f(x)= (1/2)cos(x)+sin(x) ` defined on [0,2],

For maxima / minima , find the derivative of f(x) with respect to x.

`f'(x)=(-1/2)sin(x)+cos(x)`

`(-1/2)sin(x)+cos(x)=0`

`-sin(x)+2cos(x)=0`

`-sin(x)=-2cos(x)`

`tan(x)=2`

`x=63.43^o`

`=1.107 rad`

Thus genera solution is

`x=npi+1.107`

if n=1 then `x!in[0,2]`

Thus minima/maxima lies in [0,2] if x=1.107 rad.

`f''(x)=-(1/2)cos(x)-sin(x) <0 `

`if x=1.107`

Thus f(x) has maxima at x=1.107 rad.

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