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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]`
Solution
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)`
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