# Find minimum value of...?

• August 28th 2008, 06:34 AM
fardeen_gen
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?
• August 28th 2008, 03:59 PM
mr fantastic
Quote:

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?
• August 28th 2008, 07:26 PM
fardeen_gen
I don't.
• August 28th 2008, 11:37 PM
wingless
Do you see that this function is periodic?

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

$f(x) = f(x+2\pi) ~\forall x$
• August 29th 2008, 12:02 AM
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 $2\pi$.
• August 29th 2008, 01:33 AM
Moo
Quote:

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 ? (Tongueout)

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