How to Find the Exact Location of All the Relative and Absolute Extrema of the Function f(x) = x³ – 6x² + 9x – 2?

To find the extrema of the function f(x) = x³ – 6x² + 9x – 2, we first need to determine the critical points by calculating its derivative and setting it to zero.

The first derivative of the function is:

f'(x) = 3x² - 12x + 9

Next, we set the derivative equal to zero:

3x² - 12x + 9 = 0

Factoring out 3 gives:

x² - 4x + 3 = 0

This factors to:

(x - 1)(x - 3) = 0

Thus, the critical points are:

x = 1, x = 3

To determine whether these points are relative maxima, minima, or saddle points, we can use the second derivative test. The second derivative of the function is:

f''(x) = 6x - 12

Evaluating the second derivative at the critical points:

For x = 1:

f''(1) = 6(1) - 12 = -6

Since f”(1) < 0, there is a relative maximum at x = 1.

For x = 3:

f''(3) = 6(3) - 12 = 6

Since f”(3) > 0, there is a relative minimum at x = 3.

Now, to find the absolute extrema, we also need to evaluate the function at the endpoints of the domain if it is specified, or at some reasonable interval. Here, we can choose to evaluate the function at x = 1, x = 3, and additionally consider other points that may influence the absolute maximum and minimum.

Calculating the function values:

f(1) = 1³ - 6(1)² + 9(1) - 2 = 2

f(3) = 3³ - 6(3)² + 9(3) - 2 = -2

We would also want to evaluate the function at values approaching negative or positive infinity, but typically in a constrained interval:

If we only consider these values, then:

Relative Maximum: at x = 1, f(1) = 2
Relative Minimum: at x = 3, f(3) = -2

In conclusion, the relative extrema are found at x = 1 with a value of 2 (maximum) and x = 3 with a value of -2 (minimum). To find absolute extrema, further evaluations at the endpoints or checking limits may be required based on the domain of the function.