To find the numerical derivative \( f'(2) \) using the symmetric difference quotient with \( h = 0.5 \), we can use the formula:
\[ f'(a) \approx \frac{f(a + h) - f(a - h)}{2h} \]
In this case, let \( a = 2 \) and \( h = 0.5 \). Therefore, we compute \( f(2 + 0.5) \) and \( f(2 - 0.5) \):
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Calculate \( f(2 + 0.5) = f(2.5) \): \[ f(2.5) = \frac{4}{2.5} = \frac{4 \times 10}{25} = \frac{40}{25} = 1.6 \]
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Calculate \( f(2 - 0.5) = f(1.5) \): \[ f(1.5) = \frac{4}{1.5} = \frac{4 \times 10}{15} = \frac{40}{15} = \frac{8}{3} \approx 2.6667 \]
Now we use these values in our symmetric difference quotient:
\[ f'(2) \approx \frac{f(2.5) - f(1.5)}{2h} \] \[ f'(2) \approx \frac{1.6 - \frac{8}{3}}{2 \cdot 0.5} \]
To subtract these, we need both values in terms of a common denominator. The common denominator for \( 1.6 \) (which is \( \frac{16}{10} \)) and \( 2.6667 \) (which is \( \frac{8}{3} \)) is \( 30 \):
- Convert \( 1.6 = \frac{16}{10} = \frac{48}{30} \)
- Convert \( \frac{8}{3} = \frac{80}{30} \)
Now we compute the difference:
\[ 1.6 - \frac{8}{3} = \frac{48}{30} - \frac{80}{30} = \frac{48 - 80}{30} = \frac{-32}{30} \]
Now divide by \( 2h = 1 \):
\[ f'(2) \approx \frac{-32/30}{1} = -\frac{32}{30} = -\frac{16}{15} \approx -1.0667 \]
Thus, the numerical derivative \( f'(2) \) determined using the symmetric difference quotient with \( h = 0.5 \) is approximately \(-1.0667\), or \(-\frac{16}{15}\).