Ask Questions, Get Answers

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

Verify Lagrange's law of mean for the following function;\[\]$f(x)=1-x^{2} \quad [0,3]$.

Note: This is part 1st of a 5 part question, split as 5 separate questions here.

Can you answer this question?

1 Answer

0 votes
  • Lagrange's Mean Value Theorem :
  • Let $f(x) $ be a real valued function that satisfies the following conditions.
  • (i) $f(x)$ is continuous on the closed interval $[a,b]$
  • (ii) $f(x)$ is differentiable in the open interval $(a,b)$
  • (iii) $f(a)=f(b)$
  • Then there exists atleast one value $c \in (a,b)$ such that $f'(c)=0$
Given $f(x)=1-x^2\;[0,3]$
Step 1:
$f(x)$ is a polynomial function
So it is conditions in $[0,3]$ and differentiable in $(0,3)$ the condition's for Lagrange's 
Step 2:
To find $c \in (0,3)$where $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
$\large\frac{3}{2} $$ \in (0,3)$. Lagrange's Theorem is verified


answered Jul 25, 2013 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App