24.6k views
0 votes
We have f'(x)=6x²+6x-120, so f??(x) = 12+6, which equals 0 when (x,y) = (1/2,0)

a) 12x + 6
b) 6x² + 6x - 120
c) 6x + 6
d) 12

User Jay Halani
by
8.8k points

1 Answer

1 vote

Final answer:

To determine the original function f(x) given the derivative function f'(x) = 6x² + 6x - 120, we need to integrate f'(x). After integrating each term, we find f(x) = 2x³ + 3x² - 120x + C. By substituting the given point (1/2, 0), we can find the value of C as 12. Thus, the original function is f(x) = 2x³ + 3x² - 120x + 12.

Step-by-step explanation:

The given question involves a derivative function and asks us to determine the original function. The derivative function is given as f'(x) = 6x² + 6x - 120. To find the original function f(x), we need to integrate f'(x).

We integrate each term of f'(x) to find f(x). Integrate 6x² to get 2x³, integrate 6x to get 3x², and integrate -120 to get -120x. So, f(x) = 2x³ + 3x² - 120x + C, where C is the constant of integration.

To find the value of C, we can use the given point (1/2, 0). Substitute x = 1/2 and y = 0 into the equation f(x) = 2x³ + 3x² - 120x + C and solve for C.

By substituting x = 1/2 and y = 0 into the equation, we get 0 = 2(1/2)³ + 3(1/2)² - 120(1/2) + C. Simplifying the equation gives us C = 12. Therefore, the original function f(x) is given by f(x) = 2x³ + 3x² - 120x + 12.

User Mohamed Hamza
by
7.5k points