Final answer:
To find the x-values of the relative extrema of the function f(x)=-x^3 + 3x^2 +9x + 5, we first find the critical points by setting the derivative equal to zero. Then, we use the second derivative test to determine if these critical points are relative maximum or minimum.
Step-by-step explanation:
To find the x-values of all points where the function has any relative extrema, we first need to find the critical points of the function. The critical points occur when the derivative of the function is equal to zero or undefined. In this case, the function is f(x)=-x^3 + 3x^2 +9x + 5. Taking the derivative of the function, we get f'(x)=-3x^2 + 6x + 9. Setting f'(x) equal to zero and solving for x gives us x = -1 and x = 3. These are the x-values of the critical points where the function may have relative extrema.
To determine if these critical points are relative maximum or minimum, we can use the second derivative test. Taking the second derivative of the function, we get f''(x)=-6x + 6. Evaluating f''(x) at the critical points, we find that f''(-1) = 12 and f''(3) = -12. Since f''(-1) is positive, the point x = -1 is a relative minimum. Since f''(3) is negative, the point x = 3 is a relative maximum. Therefore, the x-values of the relative extrema are -1 (relative minimum) and 3 (relative maximum).