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Find the points on the curve $y=(\cos x-1)$ in [0,2$\pi$],where the tangent is parallel to $x$-axis.

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  • 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:
Let $f(x)=\cos x-1$
Here $f(x)$ is continuous on $[0,2\pi]$ ans also differentiable at $(0,2\pi)$.
$f(0)=\cos 0-1$
$\quad\;\;\;=0$
$f(2\pi)=\cos(2\pi)-1=0$
Therefore $f(0)=f(2\pi)=0$
Hence all the conditions of Rolle's theorem are satisfied.
There exists at least one point $c\in (0,2\pi)$ for which $f'(c)=0$
Step 2:
$f(x)=\cos x-1$
$f'(x)=-\sin x$
$f'(c)=0$
$\Rightarrow -\sin c=0$
$\Rightarrow c=\pi$
Therefore $f(c)=\cos \pi-1$
$\qquad\quad\quad\;\;\;\;\;=-2$
Therefore by geometrical interpretation of Rolle's theorem $(\pi,-2)$ is a point on $y=\cos x-1$ where the tangent is parallel to $x$-axis.
answered Jul 3, 2013 by sreemathi.v
 

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