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Question
Using the Lagrange’s mean value theorem determine the values of x at which the tangent is parallel to the secant line at the end points of the given interval:
`f(x) = (x - 2)(x - 7), x ∈ [3, 11]`
Solution
f(x) = (x – 2)(x – 7), x ∈ [3, 11]
f(x) is continuous in [3, 11]
f(x) is differentiable in (3, 11)
f(3) = (3 – 2)(3 – 7) = (1)(– 4) = – 4
f(11) = (11 – 2)(11 – 7) = (9)(4) = 36
∴ f(x) is defined in the given interval.
Given that the tangent is parallel to the secant line ofthe curve between x = 3 and x = 11.
∴ f'(c) = `("f"("b") - "f"("a"))/("b" - "a")`
2c – 9 = `(36 + 4)/(11 - 3)`
where f'(x) = 2x – 9
2x – 9 = `40/8` = 5
2c = 14
⇒ c = 7 ∈ (3, 11)
∴ x = 7
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