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Find maximum value of \( f(x)=\sin x+\cos x-1 \) in \( \left[0, \frac{\pi}{2}\right] \) |
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Answer» \(f(x) = sin x + cosx - 1\; \text{in}\; [0, \frac\pi2]\) \(f'(x) = cos x - sin x\) \(\therefore f'(x) = 0\) given \(cos x - sin x = 0\) ⇒ \(tan x = 1\) ⇒ \(x = \frac{\pi}4\) \(f''(x) = - sin x - cos x \) \(f''(\frac\pi4) = -\frac{2}{\sqrt2} = -\sqrt 2 < 0\) ∴ \(x = \frac\pi4\) is point of maxima of f(x). ∴ Maximum value = \(f(\frac\pi4) = sin\frac\pi4 + cos\frac\pi 4 - 1\) \(= \frac1{\sqrt2} + \frac1{\sqrt2} - 1 \) \(=\sqrt2 -1\) |
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