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Math Help - Find minimum value of...?

  1. #1
    Super Member fardeen_gen's Avatar
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    Find minimum value of...?

    Find minimum value of 2^sin x + 2^cos x?

    How to do this? Any particular method?
    I tried finding dy/dx and setting it to zero. That gave me a relation (2^sin x)/(2^cos x) = tan x(I couldnt find out x from this). Please help?
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    Find minimum value of 2^sin x + 2^cos x?

    How to do this? Any particular method?
    I tried finding dy/dx and setting it to zero. That gave me a relation (2^sin x)/(2^cos x) = tan x(I couldnt find out x from this). Please help?
    Do you have any reason to believe an exact answer is possible?
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  3. #3
    Super Member fardeen_gen's Avatar
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    I don't.
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  4. #4
    Super Member wingless's Avatar
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    Do you see that this function is periodic?

    f(x) = 2^{\sin x} + 2^{\cos x}

    f(x) = f(x+2\pi) ~\forall x
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  5. #5
    Super Member wingless's Avatar
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    f(x) = 2^{\sin x} + 2^{\cos x}

    f'(x) = 0 ~\rightarrow~ 2^{\sin x}\cos x = 2^{\cos x}\sin x

    One obvious attempt to solve this equation is \sin x = \cos x. The solution set to this equation is x = \left \{ \frac{\pi}{4}, \frac{\pi}{4}+\pi\right \}. One of these is maximum and the other is minimum. Note that there are infinitely many maximums and minimums, as the function has a period of 2\pi.
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  6. #6
    Moo
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    Quote Originally Posted by wingless View Post
    f(x) = 2^{\sin x} + 2^{\cos x}

    f'(x) = 0 ~\rightarrow~ 2^{\sin x}\cos x = 2^{\cos x}\sin x

    One obvious attempt to solve this equation is \sin x = \cos x. The solution set to this equation is x = \left \{ \frac{\pi}{4}, \frac{\pi}{4}+\pi\right \}. One of these is maximum and the other is minimum. Note that there are infinitely many maximums and minimums, as the function has a period of 2\pi.
    Yup, these are solutions to f'(x)=0, but how can we know they're the only solutions, and maximum or minimum of the function ?

    (Ok, ok, I see... you gotta graph it.... lol)
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