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Topic: mean value theorem. 1# In each of the following functions, check the function satisfies the criteria established in Rolle's theorem and find all the values C in the given interval where F (C) =0 F(x) =x3 -4x in [-2,2]

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Answer to a math question Topic: mean value theorem. 1# In each of the following functions, check the function satisfies the criteria established in Rolle's theorem and find all the values C in the given interval where F (C) =0 F(x) =x3 -4x in [-2,2]

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Seamus
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Para verificar si la función satisface los criterios establecidos en el teorema de Rolle, necesitamos seguir estos pasos:

1. La función F(x) = x^3 - 4x es continua en el intervalo [-2, 2] ya que es un polinomio.
2. La función es derivable en el intervalo (-2, 2) ya que es un polinomio.
3. Debemos verificar si F(-2) = F(2) para asegurarnos de que se cumplan las condiciones del teorema de Rolle.

Ahora vamos a verificar si se cumple el teorema de Rolle para la función dada:

1. Calculamos F(-2) y F(2) :

F(-2) = (-2)^3 - 4(-2) = -8 + 8 = 0

F(2) = 2^3 - 4(2) = 8 - 8 = 0

2. Como F(-2) = F(2) = 0 , se cumple la condición F(a) = F(b) donde a = -2 y b = 2 .

Por lo tanto, podemos aplicar el teorema de Rolle y encontrar el valor de c en el intervalo (-2, 2) tal que F'(c) = 0 .

Calculamos la derivada de F(x) :

F'(x) = \frac{d}{dx}(x^3 - 4x) = 3x^2 - 4

Para encontrar c , igualamos F'(c) a 0:

3c^2 - 4 = 0

3c^2 = 4

c^2 = \frac{4}{3}

c = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3}

Por lo tanto, los valores de c en el intervalo [-2, 2] donde F(c) = 0 son c = -\frac{2\sqrt{3}}{3} y c = \frac{2\sqrt{3}}{3} .

\boxed{c = -\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}}

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