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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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