Question

the absolute minimum value of f(x) =to 2 sin x in[0, 3 pie/ 2] is​

Answer 1

Answer:

answer is 1

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Answer 2

Given,

$\[f\left( x \right) = 2\sin x,\left[ {0,\frac{{3\pi }}{2}} \right]\]$

Solution,

Calculate the absolute minimum value.

$\[\begin{array}{l}f\left( x \right) = 2\sin x\\ \Rightarrow f'\left( x \right) = 2\cos x\end{array}\]$

For maximum and minimum,

$\[\begin{array}{l}f'\left( x \right) = 0\\ \Rightarrow 2\cos x = 0\\ \Rightarrow \cos x = \cos \frac{\pi }{2}\\ \Rightarrow x = \frac{\pi }{2},\frac{{3\pi }}{2}\end{array}\]$

Calculate the value of function at $\[0,\frac{\pi }{2},\frac{{3\pi }}{2}\]$.

$\[\begin{array}{l}f\left( x \right) = 2\sin x\\ \Rightarrow f\left( {x = 0} \right) = 2\sin 0\\ \Rightarrow f\left( {x = 0} \right) = 0\end{array}\]$

$\[\begin{array}{l}f\left( x \right) = 2\sin x\\ \Rightarrow f\left( {x = \frac{\pi }{2}} \right) = 2\sin \frac{\pi }{2}\\ \Rightarrow f\left( {x = \frac{\pi }{2}} \right) = 2\end{array}\]$

$\[\begin{array}{l}f\left( x \right) = 2\sin x\\ \Rightarrow f\left( {x = \frac{{3\pi }}{2}} \right) = 2\sin \frac{{3\pi }}{2}\\ \Rightarrow f\left( {x = \frac{{3\pi }}{2}} \right) = - 2\end{array}\]$

Absolute minimum value is $\[ - 2\]$.