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Verify Rolle's theorem for the following function; $f(x)=\sin x, 0\leq x\leq\pi$.

Note: This is part 1st of a 4 part question, split as 4 separate questions here.

1 Answer

  • Rolle's Theorem: Let f be a real valued function that satisfies the following conditions:
  • (i) f is defined and continuous on the closed interval $[a,b]$
  • (ii) f is differentiable in the open interval $(a,b)$
  • (iii) $f(a)=f(b)$
  • Then there exists at least one value $c \in (a,b)$ such that $f'(c)=0$
$f(x)=\sin x \; 0 \leq x \leq \pi$
Step 1:
$f(x)$ is continous on $[0,\pi]$, differentiable in $(0,\pi)$ and $f(0)=f(\pi)=0$
The conditions for Rolle's Theorem are satisfied.
To find $c \in (0, \pi)$ such that $f'(c) =0$
$f'(x)=\cos x$
$f'(c)=0=> \cos c =0$ and $c=\large\frac{\pi}{2} $$\in (0, \pi)$ with $\cos {\large\frac{\pi}{2}}$$=0$
The theorem is verified
answered Jul 24, 2013 by meena.p