The value of Delta log f(x) + Delta^{2}(3^{x} | vedclass

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Question

(A) We know that $\Delta \log f(x) = \log f(x+h) - \log f(x) = \log \left[\frac{f(x+h)}{f(x)}\right]$.
Since $f(x+h) = (1 + \Delta) f(x)$,we have $\D

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$\log \left[1 + \frac{\Delta f(x)}{f(x)}\right] + 4 \cdot 3^{x}$

Vedclass · Answer

(A) We know that $\Delta \log f(x) = \log f(x+h) - \log f(x) = \log \left[\frac{f(x+h)}{f(x)}\right]$.
Since $f(x+h) = (1 + \Delta) f(x)$,we have $\Delta \log f(x) = \log \left[\frac{(1 + \Delta) f(x)}{f(x)}\right] = \log \left[1 + \frac{\Delta f(x)}{f(x)}\right]$.
Next,for the second term,$\Delta^{2}(3^{x}) = (E - 1)^{2} 3^{x} = (E^{2} - 2E + 1) 3^{x}$.
$= E^{2}(3^{x}) - 2E(3^{x}) + 3^{x} = 3^{x+2} - 2 \cdot 3^{x+1} + 3^{x}$.
$= 3^{x}(9 - 6 + 1) = 4 \cdot 3^{x}$.
Combining both terms,the result is $\log \left[1 + \frac{\Delta f(x)}{f(x)}\right] + 4 \cdot 3^{x}$.

The value of $\Delta \log f(x) + \Delta^{2}(3^{x})$ is

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