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] - Mathematics

प्रश्न

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:

$f (x) = (x - 2) (x - 7), x \in [3, 11]$

बेरीज

उत्तर

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) = $\frac{f (b) - f (a)}{b - a}$

2c – 9 = $\frac{36 + 4}{11 - 3}$

where f'(x) = 2x – 9

2x – 9 = $\frac{40}{8}$ = 5

2c = 14

⇒ c = 7 ∈ (3, 11)

∴ x = 7

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Mean Value Theorem

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 4. (ii)Q 4. (i)Q 5. (i)

पाठ 7: Applications of Differential Calculus - Exercise 7.3 [पृष्ठ २१]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board

पाठ 7 Applications of Differential Calculus
Exercise 7.3 | Q 4. (ii) | पृष्ठ २१

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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] - Mathematics

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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] - Mathematics

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