Maximum and minimum

[juantheron]

find maximum and minimum value of

[math]\displaystyle f(x) =\frac{1}{\sin x}+\frac{1}{\cos x}[/math]

where

[math]x\in \left(0,\frac{\pi}{2}\right)[/math]

[enrico___1]

Let f'(x) the first derivative of f(x)

[math]
f'(x)= -\frac{\cos x}{\sin^2 x}+\frac{\sin x}{cos^2 x}
[/math]

[math]
f'(x)= \frac{\sin^3 x-\cos^3 x}{\sin^2 x \cos^2 x}
[/math]

We impose f'(x)>0 and then we will study the critical points

[math]
\sin^3x-\cos^3x>0\to \sin x-\cos x>0\\
\sin^2x \cos^2x>0\to x\not = k\frac{\pi}{2}
[/math]

Here is the graphic of sin x >cos x (sin x: green line and cos x: black line). You can see that from 0 to

[math]\frac{\pi}{4}[/math]

(the horizontal line) sin x is greater than cos x and form

[math]\frac{\pi}{4}[/math]

to

[math]\frac{\pi}{2}[/math]

you have cos x which is greater than sin x.




There is a critical point in

[math]x=\frac{\pi}{4}[/math]

, but f(x) doesn't exists. Let's prove the continuity of f(x) in x=

[math]\frac{\pi}{4}[/math]

.

[math]
\lim_{x\to \frac{\pi}{4}^-} f(x)=\lim_{x\to \frac{\pi}{4}^+} f(x)=2\sqrt{2}
[/math]

Well you found a minimum point in x=

[math]\frac{\pi}{4}[/math]

and its value is

[math]2\sqrt{2}[/math]

[ciampax]

It's almost correct everything... except for the non existence of the function in

[math]x=\pi/4[/math]

, as the domain is

[math](0,\pi/2)[/math]

(the only two points where both the functions at denominator vanishes).