Find minimum value of...?

**fardeen_gen** · Aug 28th 2008, 06:34 AM

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?

**mr fantastic** · Aug 28th 2008, 03:59 PM

Originally Posted by fardeen_gen

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?

**fardeen_gen** · Aug 28th 2008, 07:26 PM

I don't.

**wingless** · Aug 28th 2008, 11:37 PM

Do you see that this function is periodic?

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

$f(x) = f(x+2\pi) ~\forall x$

**wingless** · Aug 29th 2008, 12:02 AM

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

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

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 $2\pi$ .

**Moo** · Aug 29th 2008, 01:33 AM

Originally Posted by wingless

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

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

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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