If f(x + y) = f(x) + f(y) + |x|y + xy^2,foral | vedclass

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(B) Given $f(x + y) = f(x) + f(y) + |x|y + xy^2$. Setting $x = 0, y = 0$,we get $f(0) = f(0) + f(0) + 0 + 0$,so $f(0) = 0$. By definition,$f'(x) = \li

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$f$ is differentiable for all $x \in R$

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(B) Given $f(x + y) = f(x) + f(y) + |x|y + xy^2$. Setting $x = 0, y = 0$,we get $f(0) = f(0) + f(0) + 0 + 0$,so $f(0) = 0$. By definition,$f'(x) = \lim_{h 	o 0} \frac{f(x + h) - f(x)}{h}$. Substituting the given functional equation: $f'(x) = \lim_{h 	o 0} \frac{f(x) + f(h) + |x|h + xh^2 - f(x)}{h}$. $f'(x) = \lim_{h 	o 0} \frac{f(h) + |x|h + xh^2}{h} = \lim_{h 	o 0} \left( \frac{f(h) - f(0)}{h} + |x| + xh ight)$. Since $f'(0) = 0$,we have $f'(x) = f'(0) + |x| + 0 = |x|$. Since $f'(x) = |x|$,the derivative exists for all $x \in R$. Thus,$f$ is differentiable for all $x \in R$.

If $f(x + y) = f(x) + f(y) + |x|y + xy^2$,$\forall x, y \in R$ and $f'(0) = 0$,then

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