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
- continuous in [a,b]
- differentiable in (a,b)
- 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]
