Question

Using the Rolle’s theorem, determine the values of x at which the tangent is parallel to the x-axis for the following functions:

`f(x) = (x^2 - 2x)/(x + 2), x ∈ [-1, 6]`

Answer 1

f(– 1) = `(1 + 2)/(-1 + 2)` = 3

f(6) = `(36 - 12)/8 = 24/8` = 3

⇒ f(– 1) = 3 = f(6)

f(x) is continuous on [– 1, 6]

f(x) is differentiable on (– 1, 6)

Now, f'(x ) = `((x + 2)(2x - 2) - (x^2 - 2x)(1))/(x + 2)^2`

= `(x^2 + 4x - 4)/(x + 2)^2`

Since the tangent is parallel to the x-axis.

f'(x) = 0

⇒ x² + 4x – 4 = 0

⇒ x = `- (4 +-  sqrt(16 + 16))/2`

x = `- (4 +-  4sqrt(2))/2`

= `- 2 +-  2sqrt(2)`

x = `- 2 +- 2sqrt(2) ∈ (-1, 6)`