Find c for Rolle's Theorem for f(x) = x3 – 12x on [0,2 3 ]

Samaj nhi aaya

bro that is not 0.2 given in question...kindly wear glass

Well, let's first check if the function f(x) = x^3 - 12x satisfies the conditions for Rolle's Theorem on the interval [0, 2^3].

Rolle's Theorem states that for a function f(x) to have a root (c) in the open interval (a, b), it must meet the following conditions:
1. f(x) must be continuous on the closed interval [a, b].
2. f(x) must be differentiable on the open interval (a, b).
3. f(a) = f(b).

In our case, the function f(x) = x^3 - 12x is indeed continuous and differentiable on the interval [0, 2^3].

Now let's check if f(0) = f(2^3).

f(0) = 0^3 - 12(0) = 0
f(2^3) = (2^3)^3 - 12(2^3) = 8^3 - 12(8) = 512 - 96 = 416

Since f(0) = 0 and f(2^3) = 416 are not equal, we can conclude that there is no value of c that satisfies the conditions of Rolle's Theorem for the function f(x) = x^3 - 12x on the interval [0, 2^3].

So, in this case, we cannot find a specific value for c. But remember, finding c is not always necessary to apply Rolle's Theorem.

To find the value of "c" that satisfies Rolle's Theorem for the function f(x) = x^3 - 12x on the interval [0, 2/3], we need to follow a few steps:

Step 1: Verify that the function satisfies the conditions for Rolle's Theorem.
For Rolle's Theorem to be applicable, we need to check if:
1. The function f(x) is continuous on the closed interval [0, 2/3].
2. The function f(x) is differentiable on the open interval (0, 2/3).

In our case, the function f(x) = x^3 - 12x is a polynomial function, which is continuous and differentiable for all real numbers.

Step 2: Verify that f(0) = f(2/3).
For Rolle's Theorem to hold, it is necessary that f(0) = f(2/3). So, let's calculate the values.

f(0) = (0)^3 - 12(0) = 0
f(2/3) = (2/3)^3 - 12(2/3) = 8/27 - 24/3 = 8/27 - 8 = 0

Since f(0) = f(2/3) = 0, it satisfies the necessary condition for Rolle's Theorem.

Step 3: Find the value of "c".
Since the function satisfies the conditions, we know that there exists a value "c" in the open interval (0, 2/3) where the derivative of the function is equal to zero.

To find "c", we need to find the derivative of f(x) and set it equal to zero, then solve for "x".

f(x) = x^3 - 12x

Taking the derivative of f(x), we get:
f'(x) = 3x^2 - 12

Setting f'(x) = 0, we have:
3x^2 - 12 = 0

Simplifying, we get:
3x^2 = 12

Dividing both sides by 3, we have:
x^2 = 4

Taking the square root of both sides, we get:
x = ±2

Since we are interested in the open interval (0, 2/3), the value of "c" that satisfies Rolle's Theorem is:
c = 2/3

First, check to make sure f(0.2) = f(3)

Assuming you meant x^3-12x,

f(3) = -9
f(0.2) = -2.4

So, Rolle's theorem does not apply here.