Browse Questions

Verify the Rolle's theorem for the function $f(x)=\sin^4x+\cos^4x\;in\;\begin{bmatrix}0,\large\frac{\pi}{2}\end{bmatrix}$

Toolbox:
• If $f[a,b]\rightarrow R$ is continuous on $[a,b]$ and differentiable on $(a,b)$ such that $f(a)=f(b)$,then there exists some c in $(a,b)$ such that $f'(c)=0$
Step 1:
$f(x)=\sin^4x+\cos^4x$ in $[0,\large\frac{\pi}{2}]$
We know that $\sin x$ and $\cos x$ is continuous everywhere and differentiable.
Therefore $\sin ^4x+\cos^4 x$ is continuous on $[0,\large\frac{\pi}{2}]$ and differentiable $[0,\large\frac{\pi}{2}]$
Also $f(0)=\sin^40+\cos^40$
$f(\large\frac{\pi}{2})=$$\sin^4\big(\large\frac{\pi}{2}\big)+$$\cos^4\big(\large\frac{\pi}{2}\big)$
$\qquad=1$
Therefore $f(x)$ satisfies conditions of Rolle's theorem on $[0,\large\frac{\pi}{2}]$,therefore there exists $c\in (0,\large\frac{\pi}{2})$
Such that $f'(c)=0$
Step 2:
Now,$f(x)=\sin^4x+\cos^4x$
$f'(x)=4\sin^3x.\cos x+4\cos^3x(-\sin x)$
$\qquad=4\sin^3x\cos x-4\cos^3x\sin x$
$\qquad=4\sin x\cos x(\sin^2x-\cos^2x)$
But $2\sin x\cos x-\sin 2x$ and $\cos^2 x-\sin^2 x=\cos 2x$
Therefore $f'(x)=2\sin 2x(-\cos 2x)$
$\qquad\qquad\;\;\;\;\;\;=-2\sin 2x.\cos 2x$
Therefore $f'(x)=0$
$\Rightarrow -2\sin 2x.\cos 2x=0$
$-2\sin 2x=0$
$\Rightarrow 2x=0$
$x=0$
Again $\cos 2x=0$
$\Rightarrow 2x=\large\frac{\pi}{2}$
$x=\large\frac{\pi}{4}$
Thus $c=0,\large\frac{\pi}{4}$ lies within the interval $[0,\large\frac{\pi}{2}]$.
Hence Rolle's theorem is verified.