#### 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) = x^{3} – 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) = x^{3} − 3x

f'(x) = 3x^{2} − 3

∴ f'(c) = 0

⇒3c^{2} − 3 = 0

⇒c^{2} − 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.