Question

Verify Lagrange’s mean value theorem for the following function:

`f(x) = x^2 - 3x - 1, x ∈ [(-11)/7 , 13/7]`.

Answer 1

The function f given as f(x) = x² – 3x – 1 is a polynomial function. Hence, it  is continuous on `[(-11)/7 , 13/7]` and differentiable on `((-11)/7, 13/7)`.

Thus, the function f satisfies the conditions of LMVT.

∴ there exists `c ∈ ((-11)/7, 13/7)` such that

f'(c) = `(f(13/7) - f((-11)/7))/(13/7 - ((-11)/7)`   ...(1)

Now, f(x) = x² – 3x – 1

∴ `f((-11)/7) = ((-11)/7)^2 - 3((-11)/7) - 1`

= `(121)/(49) + (33)/(7) - 1`

= `(121 + 231 - 49)/(49)`

= `(303)/(49)`

and `f(13/7) = (13/7)^2 - 3(13/7) - 1`

= `(169)/(49) - (39)/(7) - 1`

= `(169 - 273 - 49)/(49)`

= `(-153)/(49)`

Also, f'(x) = `d/dx(x^2 - 3x - 1)`

= 2x – 3 × 1  – 0

= 2x – 3

∴ f'(c) = 2c – 3

∴ from (1), 2c – 3 = `((-153)/49 - 303/49)/(13/7 11/7) `

∴ 2c – 3 = `-(456)/(49) xx (7)/(24)`

= `(-57)/(21)`

∴ 2c = `(-57)/(21) + 3`

= `(-57 + 63)/(21)`

= `(6)/(21)`

= `(2)/(7)`

∴ c = `(1)/(7)∈((-11)/7 , 13/7)`
Hence, Lagrange’s mean value theorem is verified.