Question

minimum value of f(x) , if f(x) = sinx in [0,2π] is . what will be the answer ?

please give me the answer of the question ❓​

Answer 1

Answer:

it is -1 at 270 degrees of sin.

Answer 2

SOLUTION

TO DETERMINE

The minimum value of f(x) , if f(x) = sin x in [0,2π]

EVALUATION

Here the given function is

$\sf f(x) = sin x$

Differentiating both sides with respect to x two times we get

$\sf f'(x) = cos x$

$\sf f'' (x) = - sin x$

For extremum value of f(x) we have

$\sf f' (x) = 0$

$\sf \implies \: cos x = 0$

$\displaystyle \sf \implies \: x = \frac{\pi}{2} \:, \: \frac{3\pi}{2}$

Now

$\displaystyle \sf f'' \bigg( \frac{\pi}{2} \bigg) = - sin \frac{\pi}{2} = - 1 < 0$

$\displaystyle \sf f'' \bigg( \frac{3\pi}{2} \bigg) = - sin \frac{3\pi}{2} = 1 > 0$

$\displaystyle \sf Thus \: f(x) \: has \: maximum \: value \: at \: x = \frac{\pi}{2} \:and \: minimum \: value \: at \: x = \frac{3\pi}{2}$

Hence the required minimum value

$\displaystyle \sf f \bigg( \frac{3\pi}{2} \bigg) = sin \frac{3\pi}{2} = - 1$

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