Question

If from Lagrange's Mean Value Theorem, we have $f\text{'}\left(x_1\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$, then .

• A. $a<x_1\le b$
• B. $a\le x_1<b$
• C. $a\le x_1\le b$
• D. $a<x_1<b$

Answer 1

The correct option is D $a<x_1<b$

According to the Lagrange's Mean Value Theorem, if f(x) is defined on [a, b] such that
(i) it is continuous on [a, b]
(ii) it is differentiable on (a, b)
Then, there exists a real number $x_1\in$(a, b) such that $f\text{'}\left(x_1\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$