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Consider the function f (x) = x³ – 7x² + 3x + 2. 1) Find the derivative, f ’(x). 2) Find the second derivative, f ’’(x). 3) Find the critical numbers of f (x) - just the x-values. 4) Use either the First Derivative Test or the Second Derivative Test to determine if these critical values are a relative maximum or relative minimum. 5) Compute the corresponding y-values and clearly label the relative maximum point and relative minimum point. 6) Find the absolute maximum point and absolute minimum point of f (x) on the interval [-2, 6]. 7) Find the inflection point of f (x).

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1) To find the derivative, f'(x), we can use the power rule. Taking the derivative of each term, we get f'(x) = 3x^2 - 14x + 3.

2) To find the second derivative, f''(x), we can differentiate f'(x). Taking the derivative of each term in f'(x), we get f''(x) = 6x - 14.

3) To find the critical numbers of f(x), we set f'(x) equal to zero and solve for x. So, 3x^2 - 14x + 3 = 0. Solving this quadratic equation, we find the critical numbers x = 1 and x = 1/3.

4) To determine if these critical values are a relative maximum or relative minimum, we can use the Second Derivative Test. Evaluating f''(x) at x = 1 and x = 1/3, we find that f''(1) = -8 and f''(1/3) = -4. Since f''(1) is negative and f''(1/3) is also negative, we can conclude that both x = 1 and x = 1/3 correspond to relative maximum points.

5) To compute the corresponding y-values, we substitute the critical numbers into the original function f(x). We find that f(1) = -1 and f(1/3) = 1/27. Therefore, the relative maximum point is (1, -1) and the other relative maximum point is (1/3, 1/27).

6) To find the absolute maximum and minimum points of f(x) on the interval [-2, 6], we evaluate f(x) at the critical numbers and the endpoints of the interval. We find that f(-2) = 36, f(6) = 218, f(1) = -1, and f(1/3) = 1/27. Therefore, the absolute maximum point is (6, 218) and the absolute minimum point is (1, -1).

7) To find the inflection point of f(x), we need to find where the concavity changes. This occurs when f''(x) = 0 or is undefined. However, in this case, f''(x) is always defined and never equals zero. Therefore, there are no inflection points in the given
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