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Thread: range

  1. #1
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    range

    Find Range of $\displaystyle f(x) = \sin^2 x-5\sin x+6$

    where $\displaystyle x\in \left[0,\pi\right]$
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  2. #2
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    Re: range

    Quote Originally Posted by jacks View Post
    Find Range of $\displaystyle f(x) = \sin^2 x-5\sin x+6$

    where $\displaystyle x\in \left[0,\pi\right]$
    $\displaystyle f'(x)=2\sin x \cdot \cos x-5 \cos x=\cos x \cdot (2\sin x-5)$

    In order to find minimum of function we have to solve following equation :

    $\displaystyle f'(x)=0 \Rightarrow \cos x=0 \Rightarrow x= \frac{\pi}{2} \Rightarrow \sin x = 1$

    So ,we can conclude that : $\displaystyle f_{min}=2$

    On the other hand function has maximum for $\displaystyle x=0$ , and $\displaystyle x=\pi$ , therefore : $\displaystyle f_{max}=6$

    So for $\displaystyle x \in [0,\pi]$ we have : $\displaystyle f(x) \in [2,6]$
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