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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)=x2-2xx+2,x∈[-1,6] - Mathematics

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प्रश्न

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]`

बेरीज

उत्तर

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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Mean Value Theorem
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Applications of Differential Calculus - Exercise 7.3 [पृष्ठ २१]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 7 Applications of Differential Calculus
Exercise 7.3 | Q 2. (ii) | पृष्ठ २१

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