Answer :

It is given that f (x) = sin x + cos x, x [0, π]

f’(x) = cosx - sinx


Now, f’(x) = 0


cosx - sinx = 0


cosx = sinx


tanx = 1


x =


Now, we evaluate the value of f at critical point and at end points of the interval [0, π]



f(0) = sin0 +cos0 = 0+1 = 1


f(π) = sin π + cos π = 0 -1 = -1


Therefore, we have the absolute maximum value of f on [0, π] is √2 occurring at


And, the absolute minimum value of f on [0, π] is -1 occurring at x = π.


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