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# Find the Value of C in Rolle'S Theorem for the Function F(X)=X3−3x in -sqrt3,0 - CBSE (Commerce) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### Question

Find the value of c in Rolle's theorem for the function f(x) = x^3 - 3x " in " (-sqrt3, 0)

#### Solution 1

f(x) = x^2 - 3x

i) f(-sqrt3) = (-sqrt3)^3 - 3(-sqrt3) = -3sqrt3 + 3sqrt3 = 0

f(0) = 0

Also  f(x) =  continuos in  [-sqrt3, 0] and differentiable in (-sqrt3,0)

f'(c) = 0

=> 3x^2 - 3 = 0

:. 3c^2 - 3 = 0

c^2 = 1

c = ±1

⇒ c = -1

#### Solution 2

The given function is f(x) = x3 – 3x.

Since a polynomial function is everywhere continous and differentiable, therefore f(x) is continous on [-sqrt3, 0] and differentaible on (-sqrt3,0)

Also f(-sqrt3) = (-sqrt3)^3 - 3(-sqrt3) = -3sqrt3 + 3sqrt3 = 0

f(0) = (0)3 – 3 × 0 = 0

Since all the three conditions of Rolle’s theorem are satisfied, so there exists a point c ∈ (-sqrt3,0) such that f'(c) = 0

f(x) = x3 − 3x

f'(x) = 3x2 − 3

∴ f'(c) = 0

⇒3c2 − 3 = 0

⇒c2 − 1 = 0

⇒ (c + 1)(c − 1) = 0

⇒ c = −1 or c = 1

Now, c != 1 [∵ 1 ∉ (-sqrt3,0)]

∴ c = -1, where c ∈ (-sqrt3,0)

Thus, the required value of c is –1.

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Solution Find the Value of C in Rolle'S Theorem for the Function F(X)=X3−3x in -sqrt3,0 Concept: Increasing and Decreasing Functions.
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