Li Hodgson
03/06/2023 · Elementary School

Si \( f(x)=x^{3}+2 x \), entonces la suma de las pendientes \( m_{t}+m_{n} \) en el punto \( x=0 \) es:

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Para encontrar la suma de las pendientes \( m_{t} + m_{n} \) en el punto \( x = 0 \) para la función \( f(x) = x^{3} + 2x \), primero necesitamos calcular la derivada de \( f(x) \). La derivada de \( f(x) \) es: \[ f'(x) = \frac{d}{dx}(x^{3} + 2x) = 3x^{2} + 2 \] Ahora, evaluamos la derivada en el punto \( x = 0 \): \[ f'(0) = 3(0)^{2} + 2 = 2 \] La pendiente de la tangente \( m_{t} \) en el punto \( x = 0 \) es \( 2 \). Si asumimos que \( m_{n} \) se refiere a la pendiente de la normal en el mismo punto, recordemos que la pendiente de la normal es el negativo del recíproco de la pendiente de la tangente. Por lo tanto: \[ m_{n} = -\frac{1}{m_{t}} = -\frac{1}{2} \] Ahora sumamos las pendientes: \[ m_{t} + m_{n} = 2 - \frac{1}{2} = 2 - 0.5 = 1.5 \] Por lo tanto, la suma de las pendientes \( m_{t} + m_{n} \) en el punto \( x = 0 \) es: \[ \boxed{1.5} \]

Quick Answer

La suma de las pendientes \( m_{t} + m_{n} \) en el punto \( x = 0 \) es 1.5.
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