Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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.

Can you answer this question?

1 Answer

0 votes
  • 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 $(-2,2)$
when $x=2$
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
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
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App