The maximum value of f(x) = x3 + sinx when x ∈ [–1, 0] is ________.

To find the maximum value of the function f(x) = x^3 + sinx when x ∈ [–1, 0], we need to follow these steps:

1. Find the derivative of the function: f'(x) = 3x^2 + cosx.

2. Set the derivative equal to zero and solve for x to find critical points: 3x^2 + cosx = 0.

3. Check the endpoints of the interval and the critical points to see where the function has its maximum value.

Let's do it:

1. The derivative of the function f(x) = x^3 + sinx is f'(x) = 3x^2 + cosx.

2. Setting the derivative equal to zero gives us 3x^2 + cosx = 0. This equation is not easy to solve analytically, but we can solve it numerically. However, in the interval x ∈ [–1, 0], cosx is always positive (0 to 1) and 3x^2 is always non-negative. Therefore, there are no solutions to the equation 3x^2 + cosx = 0 in the interval x ∈ [–1, 0].

3. Now we need to check the endpoints of the interval x ∈ [–1, 0].

   f(-1) = (-1)^3 + sin(-1) = -1 - sin(1) (approximately -1.8415)

   f(0) = (0)^3 + sin(0) = 0

Therefore, the maximum value of f(x) = x^3 + sinx when x ∈ [–1, 0] is 0.