Let fleft( frac{x + 8y}{9} right) = frac{f(x) | vedclass

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(D) Given the functional equation: $f\left( \frac{x + 8y}{9} \right) = \frac{f(x) + 8f(y)}{9}$.Differentiating both sides with respect to $x$,treating

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$9$

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(D) Given the functional equation: $f\left( \frac{x + 8y}{9} \right) = \frac{f(x) + 8f(y)}{9}$.Differentiating both sides with respect to $x$,treating $y$ as a constant:$f'\left( \frac{x + 8y}{9} \right) \cdot \frac{1}{9} = \frac{f'(x)}{9}$.This simplifies to: $f'\left( \frac{x + 8y}{9} \right) = f'(x)$.Setting $x = 0$ in the above equation:$f'\left( \frac{8y}{9} \right) = f'(0)$.Since $f'(0) = 2$,we have $f'\left( \frac{8y}{9} \right) = 2$.Let $t = \frac{8y}{9}$,then $y = \frac{9t}{8}$. Since $y$ is any real number,$t$ is also any real number. Thus,$f'(t) = 2$ for all $t$.Integrating $f'(t) = 2$ with respect to $t$ gives $f(t) = 2t + C$.Using the given condition $f(0) = -5$:$f(0) = 2(0) + C = -5 \implies C = -5$.Therefore,the function is $f(x) = 2x - 5$.To find $f(7)$:$f(7) = 2(7) - 5 = 14 - 5 = 9$.

Let $f\left( \frac{x + 8y}{9} \right) = \frac{f(x) + 8f(y)}{9}$ for all real $x$ and $y$. If $f'(0)$ exists and is equal to $2$,and $f(0) = -5$,then $f(7)$ is equal to:

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