Question

Prove rolle's theorem f(x)= sin^4x + cos^4x in[0;π/2]

Answer 1

Answer:

pls see the attachment

Answer 2

Given :

Function f ( x ) = $sin^{4} x$ + $cos^{4} x$     in  [ 0 , $\dfrac{\pi }{2}$ ]

To Prove :

The Rolle's theorem

Solution :

 f ( x ) is continuous in    [ 0 , $\dfrac{\pi }{2}$ ]

f ' ( x )  =  4 $sin^{3} x$ . cos x + 4 $cos^{3} x$ ( - sin x )

          = 4 sin x cos x ( sin² x - cos² x)      that exist in   [ 0 , $\dfrac{\pi }{2}$ ]

So, f ( x ) is differential in [ 0 , $\dfrac{\pi }{2}$ ]

Also

 f ( 0 ) = 0 + 1  = 1

and  f ' ( $\dfrac{\pi }{2}$ ) = 1 + 0 = 1

∴   f ( 0 ) =   f ' ( $\dfrac{\pi }{2}$ )

Thus, Rolle's theorem condition satisfy

Hence, There exit at least one c  ∈  [ 0 , $\dfrac{\pi }{2}$ ] such that  f ' ( c  ) = 0

∴  4 sin x cos c ( sin² c - cos² c )  = 0

   4 sin c cos c ( - cos 2 c ) = 0

Or,  - 2 sin 2 c cos 2 c = 0

Or,  -  sin 4 c = 0

Or,          4 c = π

∴                c = $\dfrac{\pi }{2}$

And     $\dfrac{\pi }{c}$  ∈ [ 0 , $\dfrac{\pi }{2}$ ]

Hence, Rolle's theorem is verified  . Answer