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e) \( f(x)=\left(x^{2}-1\right)\left(x^{2}-4\right)\left(x^{2}-9\right) \)

Ask by Gonzalez Nunez. in Argentina
Feb 19,2025

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Answer

Las raíces de la función \( f(x) = (x^2 - 1)(x^2 - 4)(x^2 - 9) \) son \( x = -3, -2, -1, 1, 2, 3 \). La función es positiva en los intervalos \( (-\infty, -3) \), \( (-1, 1) \), y \( (3, \infty) \), y negativa en \( (-3, -2) \), \( (-2, -1) \), \( (1, 2) \), y \( (2, 3) \).

Solution

Para resolver el problema, primero analizaremos la función dada: \[ f(x) = (x^2 - 1)(x^2 - 4)(x^2 - 9) \] ### Paso 1: Identificar los factores Los factores de la función son: 1. \( x^2 - 1 \) 2. \( x^2 - 4 \) 3. \( x^2 - 9 \) ### Paso 2: Encontrar las raíces de cada factor Para encontrar las raíces de cada factor, igualamos cada uno a cero: 1. \( x^2 - 1 = 0 \) \[ x^2 = 1 \implies x = \pm 1 \] 2. \( x^2 - 4 = 0 \) \[ x^2 = 4 \implies x = \pm 2 \] 3. \( x^2 - 9 = 0 \) \[ x^2 = 9 \implies x = \pm 3 \] ### Paso 3: Resumir las raíces Las raíces de la función \( f(x) \) son: - \( x = -3, -2, -1, 1, 2, 3 \) ### Paso 4: Análisis de la función La función \( f(x) \) es un polinomio de grado 6 (ya que cada factor es de grado 2 y hay 3 factores). Esto significa que la función puede tener hasta 6 intersecciones con el eje \( x \), que son las raíces que hemos encontrado. ### Paso 5: Comportamiento de la función Para entender el comportamiento de la función en los intervalos definidos por las raíces, podemos evaluar \( f(x) \) en algunos puntos de prueba en los intervalos: 1. \( (-\infty, -3) \) 2. \( (-3, -2) \) 3. \( (-2, -1) \) 4. \( (-1, 1) \) 5. \( (1, 2) \) 6. \( (2, 3) \) 7. \( (3, \infty) \) ### Paso 6: Evaluar en puntos de prueba Evaluemos \( f(x) \) en un punto de cada intervalo: 1. Para \( x = -4 \): \[ f(-4) = ((-4)^2 - 1)((-4)^2 - 4)((-4)^2 - 9) = (16 - 1)(16 - 4)(16 - 9) = 15 \cdot 12 \cdot 7 > 0 \] 2. Para \( x = -2.5 \): \[ f(-2.5) = ((-2.5)^2 - 1)((-2.5)^2 - 4)((-2.5)^2 - 9) = (6.25 - 1)(6.25 - 4)(6.25 - 9) = 5.25 \cdot 2.25 \cdot (-2.75) < 0 \] 3. Para \( x = -1.5 \): \[ f(-1.5) = ((-1.5)^2 - 1)((-1.5)^2 - 4)((-1.5)^2 - 9) = (2.25 - 1)(2.25 - 4)(2.25 - 9) = 1.25 \cdot (-1.75) \cdot (-6.75) > 0 \] 4. Para \( x = 0 \): \[ f(0) = (0^2 - 1)(0^2 - 4)(0^2 - 9) = (-1)(-4)(-9) < 0 \] 5. Para \( x = 1.5 \): \[ f(1.5) = ((1.5)^2 - 1)((1.5)^2 - 4)((1.5)^2 - 9) = (2.25 - 1)(2.25 - 4)(2.25 - 9) = 1.25 \cdot (-1.75) \cdot (-6.75) > 0 \] 6. Para \( x = 2.5 \): \[ f(2.5) = ((2.5)^2 - 1)((2.5)^2 - 4)((2.5)^2 - 9) = (6.25 - 1)(6.25 - 4)(6.25 - 9) = 5.25 \cdot 2.25 \cdot (-2.75) < 0 \] 7. Para \( x = 4 \): \[ f(4) = ((4)^2 - 1)((4)^2 - 4)((4)^2 - 9) = (16 - 1)(16 - 4)(16 - 9) = 15 \cdot 12 \cdot 7 > 0 \] ### Conclusión La función \( f(x) \) tiene raíces en \( x = -3, -2, -1, 1, 2, 3 \) y cambia de signo en cada raíz. El comportamiento de la función es positivo en los intervalos \( (-\infty, -3) \), \( (-1, 1) \), y \( (3, \infty) \), y negativo en los intervalos \( (-3, -2) \), \( (-2, -1) \), \( (1, 2) \), y \( (2, 3) \).

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Bonus Knowledge

Did you know that the function \( f(x) = (x^2 - 1)(x^2 - 4)(x^2 - 9) \) is a product of three quadratic factors? Each factor represents a parabola that opens upwards, with roots at \(-3, -2, -1, 1, 2, 3\). This means the function crosses the x-axis at these points, making these values critical for understanding its behavior and graphing it accurately. Now, let’s talk about real-world applications! Functions like this one can model a variety of real-life phenomena such as projectile motion, where different heights can be represented as the factors. Engineers might use such polynomial functions to calculate trajectories or determine points of intersection for different paths. The roots help in understanding the key locations where certain conditions might change—like when an object hits the ground!

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