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Math Help - range

  1. #1
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    range

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

    where 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 f(x) = \sin^2 x-5\sin x+6

    where x\in \left[0,\pi\right]
    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 :

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

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

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

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