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

Answer:

Option C) Rolle's theorem applies

$c = \displaystyle\frac{-4}{3}$

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

$f(x) = (x + 3)(x - 2)^2\\f'(x) = (x-2)^2 + 2(x+3)(x-2) = 3x^2-2x-8\\f'(c) = 0\\f'(c) = 3c^2-2c-8 = 0\\\Rightarrow (c-2)(3c+4) = 0\\\Rightarrow c = 2, c = \displaystyle\frac{-4}{3}$

c should lie in (-3,2)

Thus,

$c = \displaystyle\frac{-4}{3}$