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Math Help - H/W help urgent

  1. #1
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    Post H/W help urgent

    find the minimum value of \sqrt{2} sin (x - 40), hence find smallest positive value of x where the minimun value occurs.
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  2. #2
    o_O
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    You know the range for the graph y = sin(x) is [-1, 1]. For your question, you should know that the -40 inside the brackets will not affect the vertical shifting of the graph. So:

    -1 \leq \sin (x - 40) \leq 1

    -\sqrt{2} \quad \leq \quad \underbrace{\sqrt{2}\sin(x - 40)}_{\mbox{original equation}} \quad \leq \quad \sqrt{2} \quad \quad \mbox{Multiplied both sides by } \sqrt{2}

    To find what x values occur at the max and min, just set your expression equal to them
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  3. #3
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    Are you aware of stretches of translations of graphs ?

    first can you draw \sin x ? with reasonable accuracy showing where it is zero and the min/max points between -360 to 360. If you draw this graph well the rest of the problem is very easy.

    If you can do that then, you need to understand that  \sqrt{2} \sin x is \sin x stretched vertically. and that \sqrt{2} \sin (x - 40) is a horizontal shift.

    Bobak.
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  4. #4
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    Thank youuuuuuuu
    o_O and bobak

    Got the answer for the question
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