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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.

Respuesta :

Answer:

Option C) Rolle's theorem applies

[tex]c = \displaystyle\frac{-4}{3}[/tex]

Step-by-step explanation:

We are given that:

[tex]f(x) = (x + 3)(x - 2)^2[/tex]

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

[tex]f'(c) = 0[/tex]

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)

[tex]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[/tex]

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

[tex]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}[/tex]

c should lie in (-3,2)

Thus,

[tex]c = \displaystyle\frac{-4}{3}[/tex]

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