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Verify Mean Value Theorem, if \( f (x) = x^2 - 4x - 3 \) in the interval \( [a,b]\), where \( a = 1 \) and \(b = 4 \).

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Toolbox:
  • $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
$f(x)=x^2-4x-3$
It being a polynomial it is continuous in the interval [1,4] and derivable in [1,4].
$f'(x)=2x-4$
$f(4)=2(4)^2-4(4)-3$
$\qquad=16-16-3$
$\qquad=-3$
$f(1)=(1)^2-4(1)-3$
$\qquad=1-4-3$
$\qquad=-6$
Step 2:
By Mean Value Theorem we have
$f'(c)=\large\frac{f(b)-f(a)}{b-a}$
We have $f'(c)=2c-4$
$2c-4=\large\frac{-3-(-6)}{4-1}$
$2c-4=\large\frac{-3+6}{3}$
$2c-4=\large\frac{3}{3}$
$2c-4=1$
$2c=1+4$
$c=\large\frac{5}{2}$
answered May 13, 2013 by sreemathi.v
 

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