Thread: 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?

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?

I don't.

Do you see that this function is periodic?

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

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

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

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

One obvious attempt to solve this equation is $\displaystyle \sin x = \cos x$. The solution set to this equation is $\displaystyle x = \left \{ \frac{\pi}{4}, \frac{\pi}{4}+\pi\right \}$. 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 $\displaystyle 2\pi$.

Originally Posted by wingless

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

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

One obvious attempt to solve this equation is $\displaystyle \sin x = \cos x$. The solution set to this equation is $\displaystyle x = \left \{ \frac{\pi}{4}, \frac{\pi}{4}+\pi\right \}$. 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 $\displaystyle 2\pi$.

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)