Let f : [0,1] to R be such that f(xy) = f(x)f | vedclass

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(B) Given the functional equation $f(xy) = f(x)f(y)$ for all $x, y \in [0, 1].$Setting $y = 0,$ we get $f(0) = f(x)f(0).$Since $f(0) 
e 0,$ we can di

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$3$

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(B) Given the functional equation $f(xy) = f(x)f(y)$ for all $x, y \in [0, 1].$Setting $y = 0,$ we get $f(0) = f(x)f(0).$Since $f(0) 
e 0,$ we can divide by $f(0)$ to obtain $f(x) = 1$ for all $x \in [0, 1].$Now,the differential equation is $\frac{dy}{dx} = f(x) = 1.$Integrating both sides with respect to $x,$ we get $y = x + c.$Using the initial condition $y(0) = 1,$ we find $1 = 0 + c,$ so $c = 1.$Thus,the function is $y(x) = x + 1.$We need to calculate $y\left( \frac{1}{4} ight) + y\left( \frac{3}{4} ight).$$y\left( \frac{1}{4} ight) = \frac{1}{4} + 1 = \frac{5}{4}.$$y\left( \frac{3}{4} ight) = \frac{3}{4} + 1 = \frac{7}{4}.$Therefore,$y\left( \frac{1}{4} ight) + y\left( \frac{3}{4} ight) = \frac{5}{4} + \frac{7}{4} = \frac{12}{4} = 3.$

Let $f : [0,1] \to R$ be such that $f(xy) = f(x)f(y)$ for all $x, y \in [0,1],$ and $f(0) \ne 0.$ If $y = y(x)$ satisfies the differential equation $\frac{dy}{dx} = f(x)$ with $y(0) = 1,$ then $y\left( \frac{1}{4} \right) + y\left( \frac{3}{4} \right)$ is equal to

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