Question

Verify Lagrange’s mean value theorem for the following functions : f(x) = `(x - 1)/(x - 3)` on [4, 5].

Answer 1

The function f given as f(x) = `(x - 1)/(x - 3)` is a rational function which is continuous except at x = 3.
But `3 notin [4, 5]`
Hence, it is continuous on [4, 5] and differentiable on (4, 5).
Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.
∴ there exists c ∈ (4, 5) such that

f'(c) = `(f(5) - f(4))/(5 - 4)`                   ...(1)

Now, f(x) = `(x - 1)/(x - 3)`

∴ f(4) = `(4 - 1)/(4 - 3) = (3)/(1)` = 3

and f(5) = `(5 - 1)/(5 - 3) = (4)/2)` = 2

Also, f'(x) = `d/dx((x - 1)/(x - 3))`

= `((x - 3).d/dx(x - 1) - (x - 1).d/dx(x - 3))/(x - 3)^2`

= `((x - 3) xx (1 - 0) - (x - 1) xx (1 - 0))/(x - 3)^2`

= `(x - 3 - x + 1)/(x - 3)^2`

= `(-2)/(x - 3)^2`

∴ f'(c) = `(-2)/(c - 3)^2`

∴ from (1), `(-2)/(c - 3)^2`

= `(2 - 3)/(1)`
= – 1
∴ (c – 3)² = 2
∴ c – 3 = `±sqrt(2)`
∴ c = `3 ±sqrt(2)`
But `(3 - sqrt(2)) notin (4, 5)`
∴ c ≠ `3 - sqrt(2)`
∴ c = `3 + sqrt(2) ∈ (4, 5)`
Hence, Lagrange’s mean value theorem is verified.