Let f: R to R be such that f(xy) = f(x)f(y) f | vedclass

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(C) Given $f(xy) = f(x)f(y)$ and $f(0) 
e 0$. Since $f(0) = f(0 \cdot x) = f(0)f(x)$, and $f(0) 
e 0$, we have $f(x) = 1$ for all $x \in R$.Substitu

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$15$/$32$

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(C) Given $f(xy) = f(x)f(y)$ and $f(0) 
e 0$. Since $f(0) = f(0 \cdot x) = f(0)f(x)$, and $f(0) 
e 0$, we have $f(x) = 1$ for all $x \in R$.Substituting $f(x) = 1$ into the given integral equation: $x^2 g(x) = \int_1^x (t^2 - t g(t)) dt$.Differentiating both sides with respect to $x$ using the Leibniz rule: $2x g(x) + x^2 g'(x) = x^2 - x g(x)$.Rearranging the terms: $x^2 g'(x) + 3x g(x) = x^2$.Dividing by $x^2$ (for $x \ge 1$): $g'(x) + \frac{3}{x} g(x) = 1$.This is a linear differential equation of the form $g'(x) + P(x)g(x) = Q(x)$, where $P(x) = 3/x$ and $Q(x) = 1$.The integrating factor is $IF = e^{\int (3/x) dx} = e^{3 \ln x} = x^3$.The general solution is $g(x) \cdot x^3 = \int (1 \cdot x^3) dx = \frac{x^4}{4} + C$.At $x = 1$, the integral $\int_1^1 (t^2 - t g(t)) dt = 0$, so $1^2 g(1) = 0 \implies g(1) = 0$.Substituting $x = 1$ into the solution: $0 \cdot 1^3 = \frac{1^4}{4} + C \implies C = -1/4$.Thus, $g(x) x^3 = \frac{x^4 - 1}{4}$, which gives $g(x) = \frac{x^4 - 1}{4x^3}$.For $x = 2$, $g(2) = \frac{2^4 - 1}{4(2^3)} = \frac{16 - 1}{4(8)} = \frac{15}{32}$.

Let $f: R \to R$ be such that $f(xy) = f(x)f(y)$ for all $x, y \in R$ and $f(0) \ne 0$. Let $g: [1, \infty) \to R$ be a differentiable function such that $x^2 g(x) = \int_1^x (t^2 f(t) - t g(t)) dt$. Then $g(2)$ is equal to:

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