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Supongamos que la función \( h \) está definida para los números reales de la siguiente manera. \[ \begin{array}{ll}2 & \text { si } x<-2 \\ -(x+1)^{2}+3 & \text { si }-2 \leq x<1 \\ \frac{1}{4} x-2 & \text { si } x \geq 1\end{array} \] Calcular \( h(-5), h(-2) \) y \( h(0) \) \( h(-5)=\square \) \( h(0)=\square \)

Ask by Frazier Pierce. in Mexico
Mar 17,2025

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\( h(-5) = 2 \), \( h(-2) = 2 \), \( h(0) = 2 \)

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The Deep Dive

Para calcular \( h(-5) \) y \( h(0) \) usando la definición de la función \( h \): 1. Para \( h(-5) \): Dado que \( -5 < -2 \), utilizamos la primera parte de la función: \[ h(-5) = 2. \] 2. Para \( h(-2) \): Aquí, \( -2 \) se encuentra en la segunda parte de la función, que se aplica cuando \( -2 \leq x < 1 \): \[ h(-2) = -((-2) + 1)^{2} + 3 = -(-1)^{2} + 3 = -1 + 3 = 2. \] 3. Para \( h(0) \): Como \( 0 \) también se encuentra en la segunda parte de la función, calculamos: \[ h(0) = -((0) + 1)^{2} + 3 = -(1^{2}) + 3 = -1 + 3 = 2. \] Por lo tanto: \[ h(-5) = 2 \] \[ h(-2) = 2 \] \[ h(0) = 2 \]

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