Given f''(x) = 4x - 2 and f'(-2) = 3 and f(-2) = -3. Find f'(x) and find f(3), please!

Note that $f''(x) = \dfrac{d}{dx} f'(x)$.

Then $\dfrac{d}{dx} f'(x) = 4x - 2$.

Integrate on both sides, we have $f'(x) = 2x^2 - 2x + C$ for some real constant C. (Exercise: Try to write out the integration process on your own.)

Since f'(-2) = 3, we know that

$2(-2)^2 - 2(-2) + C = 3\\ C = -9$

Then f'(x) = 2x^2 - 2x - 9.

To find f(x), it is pretty much the same process, just with f''(x) replaced with f'(x) and f'(x) replaced with f(x). Please try to do that on your own.