Question

Determine whether Rolle's Theorem can be applied to the function f(x) = (x + 3)(x - 2)^2 on the closed interval {-3, 2}. If Rolle's Theorem can be applied, find all numbers c in the open interval (-3,2) such that f'(c) = 0.

Rolle's Theorem applies, -(8/3)

Rolle's Theorem applies, -(4/5)

Rolle's Theorem applies, -(4/3)

Rolle's Theorem does not apply.

Answer 1

Rolle's Theorem can be applied to the function f(x) = (x + 3)(x - 2)^2 on the closed interval {-3, 2}, and the number c in the open interval (-3, 2) where f'(c) = 0 is c = -4/3.

Step-by-step explanation:

To determine whether Rolle's Theorem can be applied to the function f(x) = (x + 3)(x - 2)^2 on the closed interval {-3, 2}, we need to check if the function satisfies the three conditions for Rolle's Theorem:

1. f(x) is continuous on the closed interval [-3, 2].
2. f(x) is differentiable on the open interval (-3, 2).
3. f(-3) = f(2).

First, note that the function f(x) = (x + 3)(x - 2)^2 is a polynomial function, and all polynomial functions are continuous and differentiable for all real numbers. Therefore, condition 1 is satisfied. Next, we find the derivative of f(x):

f'(x) = (2x-4)(x-2)+(x+3)(2(x-2)) = 4x^2 - 12x + 8 + 2x^2 - 2x - 12 = 6x^2 - 14x - 4.

To find the values of c in the open interval (-3, 2) where f'(c) = 0, we set f'(x) = 0:

6x^2 - 14x - 4 = 0.

We can solve this quadratic equation using factoring, the quadratic formula, or by completing the square. Solving the equation, we find the roots as x = -4/3 and x = 2/3. However, only the root x = -4/3 is in the interval (-3, 2). So, Rolle's Theorem can be applied, and the number c in the open interval (-3, 2) where f'(c) = 0 is c = -4/3.

Answer 2

Option C) Rolle's theorem applies

`c = \displaystyle(-4)/(3)`

Step-by-step explanation:

We are given that:

`f(x) = (x + 3)(x - 2)^2`

Closed interval: [-3,2]

Rolle's Theorem:

According to this theorem if the given function

1. continuous in [a,b]
2. differentiable in (a,b)
3. f(a) = f(b)

the, there exist c in (a,b) such that

`f'(c) = 0`

Continuity of function:

Since the given function is a continuous function, it is continuous everywhere. Therefore, f(x) is continuous in [-3,2]

Differentiability of function:

A polynomial function is differentiable For all arguments. Therefore, f(x) is differentiable in (-3,2)

Now, we evaluate f(-3) and f(2)

`f(x) = (x + 3)(x - 2)^2\\f(-3) = (-3+3)(-3-2)^2 = 0\\f(2) - (2+3)(2-2)^2 = 0\\\Rightarrow f(-3) = f(2) = 0`

Thus, Rolle's theorem applies on the given function f(x).

According to Rolle's theorem there exist c in (a,b) such that f'(c) = 0

`f(x) = (x + 3)(x - 2)^2\\f'(x) = (x-2)^2 + 2(x+3)(x-2) = 3x^2-2x-8\\f'(c) = 0\\f'(c) = 3c^2-2c-8 = 0\\\Rightarrow (c-2)(3c+4) = 0\\\Rightarrow c = 2, c = \displaystyle(-4)/(3)`

c should lie in (-3,2)

Thus,

`c = \displaystyle(-4)/(3)`