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The $'c'$ of Lagranges Mean Value Theorem for the function $f(x)=x^{2}+2x-1;a=0,b=1$ is

\[\begin{array}{1 1}(1)-1&(2)1\\(3)0&(4)\frac{1}{2}\end{array}\]

1 Answer

$f(x)= x^2+2x-1$
$f(0)=-1$
$f(1)= 1+2-1=2$
$f'(x) =2x+2$
$f'(c) =2c+2$
$f'(c) =\large\frac{f(1) -f(0) }{1-0}$
$2c+2= \large\frac{2-(-1) }{1-0}$
$2c+2=2+1=3$
$2c=1$
$c= \large\frac{1}{2}$
Hence 4 is the correct answer.
answered May 19, 2014 by meena.p
 
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