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Using Rolle's theorem,find the point on the curve $y=x(x-4),x\in[0,4]$,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 $f(x)=x(x-4)$
Clearly $f(x)$ is a polynomial function,which is continuous on $[0,4]$ and differentiable on $(0,4)$
$f(0)=0(0-4)$
$\qquad=0$
$f(4)=4(4-4)$
$\qquad=0$
Therefore $f(0)=f(4)=0$
Thus all the conditions of Rolle's theorem is verified.Consequently there exists at least one point $c\in (0,4)$ for which $f'(c)=0$
But $f'(c)=0$
$f(x)=x(x-4)=x^2-4x$
Step 2:
Therefore $f'(x)=2x-4$
$f'(c)=0\Rightarrow 2c-4=0$
$\Rightarrow 2c=4$
$c=2$
Therefore $f(c)=2(2-4)$
$\qquad\qquad\;\;\;\;\;=-4$
By geometrical interpretation of Rolle's theorem $(-4,2)$ is the point on $y=x(x-4)$,where tangent is parallel to $x$-axis.
answered Jul 3, 2013 by sreemathi.v
 

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