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Verify mean value theorem for each of the function $f(x)=x^3-2x^2-x+3\;in\;[0,1].$

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Toolbox:
  • Let $f(x)$ be a function defined on $[a,b]$ such that
  • (i) it is continuous on $[a,b]$
  • (ii) it is differentiable on $(a,b)$
  • Then,there exists $c\in (a,b)$ such that $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
$f(x)=x^3-2x^2-x+3$ in $[0,1]$
Since $f(x)$ is a polynomial function it is continuous everywhere in $[0,1]$
It is also differentiable on $[0,1]$
Hence both the conditions of Lagrange's Mean value theorem are satisfied.
So there must be one real number $c\in (0,1)$ such that
$f'(c)=\large\frac{f(1)-f(0)}{1-0}$
$f(1)=1-2-1+3=1$
$f(0)=3$
Step 2:
$f'(x)=3x^2-4x-1$
$\Rightarrow f'(c)=3c^2-4c-1$
Therefore $3c^2-4c-1=\large\frac{1-3}{1}$
$\Rightarrow 3c^2-4c-1=-2$
$\Rightarrow 3c^2-4c-1=0$
On solving for $c$ we get,
$3c^2-3c-c+1=0$
$3c(c-1)-1(c-1)=0$
$(3c-1)(c-1)=0$
$\Rightarrow c=\large\frac{1}{3}$$,1$
Clearly $(\large\frac{1}{3},$$1)$ lies on $[0,1]$
Hence Lagrange's mean value theorem is verified.
answered Jul 3, 2013 by sreemathi.v
 
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