Function with the property $f(a+b)=f(a)+f(b)+2ab$

Question

Let $f: \mathbb{R} \to \mathbb{R}$ and let's define $f$ such that the following property holds $$f(a+b)=f(a)+f(b)+2ab \text{ for all } a,b.$$ Also let the function be differentiable at $0$ and $f'(0) = 3.$

Show that the function is differentiable everywhere and determine the derivative $f'(x)$.

Since $f(0) = 0$ and we have that $\lim_{h\to0} \frac{f(h)}{h} = 3$.

So $$\frac{f(x+h)-f(h)}{h} = \frac{f(x)+f(h)+2xh}{h} = \frac{f(h)}{h}+2xh = 3+2xh$$

so it's differentiable and the derivative is $f'(x)=3+2xh$? I feel like i should have gotten rid of the $h$?

Answer 1

You almost get it correctly, except that the definition of $f'(x)$ should be:

$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

So, the argument should look like this: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to0}\frac{(f(x)+f(h)+2xh)-f(x)}{h} = \lim_{h\to0}\left(\frac{f(h)}h + 2x\right)$$

Answer 2

There is a tiny mistake:

$$f'(x) = \lim_{h\to 0}\frac{f(x+h) - f(\color{red}x)}{h} = \lim_{h\to 0}\frac{f(h) + 2xh}{h} = f'(0) + 2x = 3+2x.$$

PS. From there,

$$f(x) = f(0) +x f'(0) + x^2 = 3x + x^2.$$

Answer 3

We could also solve this question with partial derivatives. I find this method to be quite straightforward and easy to evaluate: $$f(a+b)=f(a)+f(b)+2ab$$ Partially differentiating w.r.t. a: $$f'(a+b)=f'(a)+2b$$ Substituting $a=0$ because we know $f'(0)=3$ : $$ f'(b)=3+2b$$ Solving this differential equation: $$f(b)=3b+2b^2+c$$ Since $f(0)=0$ , $c$ is $0$, giving us: $$\boxed {f(x)=3x+2x^2}$$

Answer 4

HINT: Note that $$f(a+b)-(a+b)^2=(f(a)-a^2)+(f(b)-b^2)$$ any use Cauchy functional equation.