<s>[INST] 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.
 [/INST] 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!
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