Function question $f(x+1)$ and $f(x)$ 
Given that $f(x+1)-f(x)=4x+5$, $f(0)=6$. Find $f(x)$.

My attempt, 
$f(1)-6=5$, 
$f(1)=11$
How to proceed then? I've never solve this kind of question before.
 A: This type of question is called a functional equation. First, suppose $f(x)$ is in the form
$$f(x)=ax+b$$
Then you can use this assumption to solve for $a$ and $b$. So
$$a(x+1)+b-ax-b=4x+5$$
$$a=4x+5$$
Which cannot be, since $a$ is a constant. So $f(x)$ must not be of that form. So, instead consider the quadratic form
$$f(x)=ax^2+bx+c$$
Then we have
$$a(x+1)^2+b(x+1)+c-ax^2-bx-c=4x+5$$
$$a(x^2+2x+1)+b(x+1)-ax^2-bx=4x+5$$
$$ax^2+2ax+a+bx+b-ax^2-bx=4x+5$$
$$2ax+a+b=4x+5$$
So now we have a system that we can use to solve for the constants $a$ and $b$:
$$2a=4$$
$$a+b=5$$
Which gives us $a=2$, $b=3$. Now you need to solve for $c$ given that $f(0)=6$, giving us $c=6$, so
$$f(x)=2x^2+3x+6$$
If you have any questions, just ask!
A: Without further assumptions, there are infinite solutions. You may define $f(x)$ over the interval $(0,1)$ in any way you like, then set $f(0)=6$. The recurrence $f(x+1)=4x+5+f(x)$ allows to define $f$ on $[1,2)$, then on $[2,3)$ and so on. The same recurrence written as $f(x)=f(x+1)-(4x+5)$ allows to define $f$ on $[-1,0)$, then on $[-2,-1)$ and so on.
$$f(x) = 2x^2+3x+6+2017 \sin(2\pi x) $$
is an example of a continuous, non-polynomial solution and
$$ f(x) = 2x^2+4x+6-\left\lfloor x\right\rfloor $$
is an example of a discontinuous solution. It is true that
$$ \forall x\in\mathbb{Z},\qquad f(x)=2x^2+3x+6 $$
as shown in other answers, but that is not enough to fix the values of $f$ over $\mathbb{R}$. If $f(x)$ is a solution and $g(x)$ is a $1$-periodic function such that $g(0)=0$, $f(x)+g(x)$ is still a solution.
A: You can telescope it!
For $x=1$ you have $f(2)-f(1)=9$
For $x=2$ you have $f(3)-f(2)=13$
For $x=3$ you have $f(4)-f(3)=17$
For $x=4$ you have $f(5)-f(4)=21$
...
For $x=k$ you have $f(k+1)-f(k)=4k+5$
If you add all of this together, you get
$-f(1)+f(k+1)=9+13+17+21+\cdots+(4k+5)$
Or
$f(k+1)=f(1)+\sum_{i=1}^k (4i+5)$
