Find minimum value of...?

fardeen_gen · #1 $Old$ 08-28-2008, 08:34 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 · #2 $Old$ 08-28-2008, 05:59 PM

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?

fardeen_gen · #3 $Old$ 08-28-2008, 09:26 PM

I don't.

wingless · #4 $Old$ 08-29-2008, 01:37 AM

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 · #5 $Old$ 08-29-2008, 02:02 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 · #6 $Old$ 08-29-2008, 03:33 AM

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)