Hfrhyu

Given that $forall x,y in mathbbR, f(x+y)=f(x)+xcdot y + y$, determine $f'(x)$.

I tried plugging in spacial cases like letting $x=y=0$, but it did not seem to get me anywhere.

(The question does not state anything about continuity. What should I assume in such cases?)

Answer: $f'$ does not exist.

If we put $y=-x$ we get $$f(0) = f(x)-x^2-x,$$ so $$f(x)=x^2+x+a$$ where $a=f(0).$

So plugging this into the starting equation we get $$(x+y)^2+x+y+a = x^2+x+a+xy+y$$

and we get $$y(x+y) =0;;;forall x,y,$$

a contradiction. Such a function does not exist, so $f'(x)$ also does not exist.

Regarding continuity, we could use the fact that the functional equation $$f(x+y)=f(x)+xcdot y+ytag1$$ holds for all real $x,y$ to prove that $f$ must be continuous everywhere.

First of all, it should be clear from $(1)$ that $f$ must be defined on the whole real line. Such a function $f$ will be continuous if and only if for all $epsilon>0$ and all $xinBbb R,$ there is some $delta>0$ such that $$bigl|f(x+y)-f(x)bigr|<epsilon$$ for all $yin(-delta,delta).$ Since begineqnarraybigl|f(x+y)-f(x)bigr| &overset(1)=& bigl|f(x)+xcdot y+y-f(x)bigr|\ &=& |xcdot y+y|\ &=& bigl|(x+1)cdot ybigr|\ &=& |x+1|cdot|y|,endeqnarray this means we must show that for all $epsilon>0$ and $xinBbb R,$ there is some $delta>0$ such that $|x+1|cdot|y|<epsilon$ whenever $|y|<delta.$ Would you be able to take it from there?

However, it would be even more straightforward to prove that $f$ is differentiable everywhere (by explicitly calculating the derivative), since differentiable functions are continuous. begineqnarrayf'(x) &:=& lim_hto 0fracf(x+h)-f(x)h\ &overset(1)=& lim_hto 0fracf(x)+xcdot h+h-f(x)h\ &=& lim_hto 0frac(x+1)cdot hh\ &=& lim_hto 0x+1\ &=& x+1.endeqnarray

As a result, we can say that there is some real $c$ such that $$f(x)=frac12x^2+x+ctag2$$ for all real $x.$ Our job, then, is to determine what value(s) of $c$ (if any) will allow $f$ to satisfy $(1)$ for all real $x,y.$

We can see that we need begineqnarrayf(x)+xcdot y+y &overset(1)=& f(x+y)\ &=& frac12(x+y)^2+(x+y)+c\ &=& frac12left(x^2+2xcdot y+y^2right)+x+y+c\ &=& frac12x^2+xcdot y+frac12y^2+x+y+c\ &=& frac12x^2+x+c+xcdot y+y+frac12y^2\ &overset(2)=& f(x)+xcdot y+y+frac12y^2endeqnarray for all real $x,y.$ Unfortunately, this only holds when $y=0,$ so no value of $c$ will work. Thus, $f$ does not exist, nor does $f'$.

Hint $:$ For all $h neq 0$ observe that $frac f(x+h) - f(x) h = x + 1.$ So $$f'(x)=limlimits_h rightarrow 0 frac f(x+h) - f(x) h = x+1.$$