If $$f(x+y)$$−$$f(x)$$−$$f(y)=$$|x|$$y+xy2$$∀x,y∈R and f′$$(0)=0$$ then

$$f1(x)=Lx$$→$$0f(x+h)$$−$$f(x)h=Lh$$→$$0f(h)+$$|x|$$h+xh2h=Lth$$→$$0$$⁡$$f1(h)+$$|x|$$+xh=$$|x|

Differentiability

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Question

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

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Solution

$\begin{array}{l} f^{1} (x) = L_{x \to 0} \frac{f (x + h) - f (x)}{h} = L_{h \to 0} \frac{f (h) + | x | h + x h^{2}}{h} \\ = {Lt}_{h \to 0} (f^{1} (h) + | x | + x h) = | x | \end{array}$

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