Lagranges mean value theorem
If a function in continuous in [a,b] and differentiable in (a,b) given that a not equals b .Then curve will have slope at each and every point in (a,b)
The slope of the curve is given by \[f'(c)=\frac{f(b)-f(a)}{b-a}\]
c lies between a and b
Example - verify Lagranges Mean value theorem for \[f(x)=x^2+2x+3\] in [4,6]
solution \[f(x)=x^2+2x+3\] in [4,6]
condition 1 It is continuous in [4,6]
condition 2 derivable in (a,b)
both the conditions are satisfied so there exist at least one real number c in (4,6) such that
\[f'(c)=\frac{f(6)-f(4)}{6-4}\]=27 ..............................(1)
\[f'(x)=2x+2\].......................................(2)
equating equation 1 and 2 we get c =5 which lies between (4,6)
hence Lagranges theorem is verified and c =5
