Let f (x) be a function that f ′(x) = x2 + 2x + 2. Suppose that f (−1) = −1. What is f (2)?A. -1B. 1C. 11D. 17E. -17

To find f(2), we first need to find the antiderivative of f'(x) = x^2 + 2x + 2.

The antiderivative of x^2 is (1/3)x^3, the antiderivative of 2x is x^2, and the antiderivative of 2 is 2x. So, the antiderivative of f'(x) is (1/3)x^3 + x^2 + 2x + C, where C is the constant of integration.

We know that f(-1) = -1, so we can plug these values into our antiderivative to solve for C:

-1 = (1/3)(-1)^3 + (-1)^2 + 2(-1) + C -1 = -1/3 + 1 - 2 + C -1 = -4/3 + C C = -1 + 4/3 = 1/3

So, our function f(x) is (1/3)x^3 + x^2 + 2x + 1/3.

Finally, we can find f(2) by plugging 2 into our function:

f(2) = (1/3)(2)^3 + (2)^2 + 2(2) + 1/3 f(2) = 8/3 + 4 + 4 + 1/3 f(2) = 17

So, the answer is D. 17.