question_answer

Suppose \[f(x)=a{{x}^{2}}+b{{x}^{2}}+cx+d,\] \[x\in [0,\,1]\] is continuous in given closed interval and differential in open interval. Then, verify the Lagrange?s mean value theorem.

Answer:

\[f(x)=a{{x}^{2}}+b{{x}^{2}}+cx+d,\] \[x\in [0,\,1]\] Here, we see that first two conditions of Lagrange's mean value theorem is already exist in the question. Thus, their exist at least one number c such that \[0<c\text{ }<1\]and \[0<\frac{f(b)-f(a)}{b-a}=f'(c')\] \[\Rightarrow \]   \[\frac{f(1)-f(0)}{1-0}=2ac'+2bc'+c\] \[[\because \,\,\,f'(x)=2ax+2bx+c]\] \[\Rightarrow \]   \[\frac{a+b+c+d-d}{1}=2c'(a\,+b)+c\] \[\Rightarrow \]   \[a\,+b=2c'(a\,+b)\] \[\Rightarrow \]   \[c'=\frac{1}{2}\in (0,\,1)\] Hence, Lagrange's mean value is satisfied.