If from Lagrange’s mean value theorem, we have (f'(x) = f'(b) - f(a))/b - a, then A. a < x1 ≤ b

Question

If from Lagrange’s mean value theorem, we have \(\frac{f'(x) = f'(b) - f(a)}{b - a},\) then

A. a < x₁ ≤ b

B. a ≤ x₁ < b

C. a < x₁ < b

D. a ≤ x₁ ≤ b

Answer 1

Correct answer is C.

\(\text {f'(x)} = \frac{\text{f(b) - f(a)}}{\text{b - a}}\)

In the Lagrange’s Mean Value Theorem, c Є (a, b) such that \(\text {f'(c)} = \frac{\text{f(b) - f(a)}}{\text{b - a}}\)

f is continuous on [a, b] and differentiable on (a, b), then there exists a real number x Є (a, b)

So, in the case of x₁, f'(x₁) \( = \frac{\text{f(b) - f(a)}}{\text{b - a}},\) then x₁Є (a, b)

\(\therefore\) a < x₁ < b