Question

Given f''(x)=6x-3 and f'(-2)=-4 and f(-2)=5, what is f'(x)? And what is f(2)?

A. f'(x) = 3x² - 3x + C and f(2) = -3
B. f'(x) = 3x² - 3x + 1 and f(2) = -3
C. f'(x) = 3x² - 3x + C and f(2) = -1
D. f'(x) = 3x² - 3x + 1 and f(2) = -1

Answer 1

To find f'(x), integrate f''(x) and get 3x^2 - 3x + C. To find f(2), integrate f'(x) and get x^3 - 3/2x^2 + Cx + D, then substitute x = 2.

Step-by-step explanation:

To find f'(x), we can integrate f''(x) with respect to x. Since f''(x) = 6x - 3, integrating once gives us f'(x) = 3x^2 - 3x + C, where C is a constant of integration.

To find f(2), we can integrate f'(x) with respect to x. From the previous result, we have f'(x) = 3x^2 - 3x + C. Integrating once more gives us f(x) = x^3 - rac{3}{2}x^2 + Cx + D, where D is another constant of integration. Plugging in x = 2 into this expression gives us f(2) = 8 - 6 + 2C + D.