Question

Maths , Continuity and differentiability 3 marks

Answer 1

Lagrange's Mean Value Theorem states that if f(x) is continuous on [a,b] and differentiable on (a,b), then there exists c ∈ (a,b) such that

$\frac{f(b) - f(a)}{b - a} = f'(c).$

For the function f(x) = log(x), on the interval [1, 2], we have a = 1 and b = 2.

So,  

$\frac{f(b) - f(a)}{b - a} = \frac{\log(2) - \log(1)}{2-1} = \frac{\log(2)}{1} = \log(2).$

Now, f'(x) = 1/x and so f'(1 / log(2)) = log(2).

So if we set c = 1 / log(2), then Lagrange's Mean Value Theorem is verified for this f(x), as required.