Let $f:\mathbb R\to\mathbb R$ be a differentiable function such that $2f(x+y)+f(x-y)=3f(x)+3f(y)+2xy \;\forall x,y\in\mathbb R$ and $f'(0)=0$. Find the value of $f(5)+f'(5)$.
Solution:
Putting $x=y=0, f'(0)=0$
$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{f(x)-f(x-h)+3f(h)+2xh}{2h}$
$=\lim_{h\to0}\frac{f(x)-f(x-h)}{2h}+\frac32\lim_{h\to0}\frac{f(h)}h+x$
$=\frac12f'(x)+x\implies f'(x)=2x\implies f(x)=x^2$
$\therefore f(5)+f'(5)=25+10=35$
My doubt:
I think when they expanded $f(x+h)$, it should have been $2f(x)$ instead of $f(x)$, i.e.,
$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{2f(x)-f(x-h)+3f(h)+2xh}{2h}$
But now I don't know what to do with the extra $f(x)$.
Best Answer
The solution is correct. Since $$2f(x+h)=3f(x)-f(x-h)+3f(h)+2xh,$$ we have \begin{align*} f(x+h)-f(x)&=\frac{2f(x+h)-\color{red}{2}f(x)}{2}\\ &=\frac{3f(x)-f(x-h)+3f(h)+2xh-\color{red}{2}f(x)}{2}\\ &=\frac{f(x)-f(x-h)+3f(h)+2xh}{2}. \end{align*}