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Using Rolle's theorem find the points on the curve $y=x^{2}+1,-2\leq x \leq 2$ where the tangent is parallel to $x$- axis.

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Toolbox:
  • 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 $f'(c)=0$
$y=x^2+1\; for \;x \in [-2,2]$
Step 1:
y is a polynomial function it is continuous in $[-2,2]$ and differentiable is $(-2,2)$
$y=4+1=5$
when $x=2$
$y=4+1=5$
For $f(x)=y \qquad f(-2)=f(2)=5$
The conditions for Roller's theorem are satisfied.
Step 2:
There is atleast one value $c \;\in\; (-2,2)$
Where $\large\frac{dy}{dx}$$=0$ At this point the slope of the tangent is 0
$\therefore $ it will be $||$ to the x-axis
$\large\frac{dy}{dx}$$=0=>2x=0=>x=0$
When $x=0\qquad y=1$
The slope of the tangent at $(0,1)$ is parallel to the x-axis
answered Jul 24, 2013 by meena.p
 
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