Approximate $f(1.01)$ for a function satisfying $f'(x) = 3f(x) + 3x$ and $f(1)=3$ Suppose that the derivative of a function satisfies the formula $f'(x) = 3f(x) + 3x$. If  $f(1)=3$, use linear approximation to estimate the value of the function at $1.01$.
I think I found 
$f(x) = 3x$.
I do not need the answer (you can still provide it if you want), I just need to know how to do this. I kept reading my book but I do not know what steps to take to tackle this questions. Maybe someone is able to explain to me clearly.
 A: Linear approximation is nothing more then the tangent line to the graph, that is: $$f(x) \approx f(x_0)+f'(x_0)(x-x_0).$$
Other way of writing this is: $$f(x_0+h) \approx f(x_0) + f'(x_0) \cdot h.$$
The issue here is the following: you don't have an explicit expression for $f$, so, a priori, we can't find a general expression for $f'(x)$. Then comes in the relation that you were given: $$f'(x) = 3f(x)+3x.$$
We can solve the differential equation to find the expression for $f$, but let's leave that for another day. Evaluating the relation given at $1$, we have: $$f'(1) = 3f(1) + 3\cdot 1 = 12.$$
This way: $$f(1.01) \approx f(1)+f'(1) \cdot 0.01 = 3+12\cdot 0.01 = 3.12.$$
Ok?
A: Use the Taylor approximation:
$$f(a+h)\approx f(a)+f'(a)h$$
for $a=1$ and $h=0.01$. In your case, $f'(1)$ can be expressed with your formula:
$$f(1+h)\approx f(1)+(3f(1)+3\cdot 1)h$$
you have everything you need, just plug it in.
So, they don't actually expect you to solve the differential equation. But you can, it's a simple linear differential equation with constant coefficient. The general solution is
$$f(x)=Ce^{3x}-x-\frac13$$
for arbitrary $C$. From the condition $f(1)=3$ you get $C=13/(3e)$. However, as I said, they don't expect you to do this.
