How do you find the exact relative maximum and minimum of the polynomial function of #f(x) = x^3 - 3x^2 - x + 1#?

Question

Ezra Adams · Accepted Answer

The first derivative finds the inflection points. The second derivative indicates whether it is a maximum or minimum.

Suppose f(x) is a function of x that is twice differentiable at a stationary point x_0.
1. If #f''(x_0)>0#, then f has a local minimum at x_0.
2. If #f''(x_0)<0#, then f has a local maximum at x_0.

Weisstein, Eric W. "First Derivative Test." From MathWorld--A Wolfram Web Resource. https://tutor.hix.ai
Weisstein, Eric W. "Second Derivative Test." From MathWorld--A Wolfram Web Resource. https://tutor.hix.ai

Inflection points at:
#f(x) = x^3 − 3x^2 − x + 1#
#f'(x) = 3x^2 − 6x -1 = 0#

Local min/max from:
#f''(x) = 6x − 6# at the inflection points.

Axel Alvarez · Answer

undefined To find the exact relative maximum and minimum of the polynomial function $ f(x) = x^3 - 3x^2 - x + 1 $, you can follow these steps:

1. Find the critical points by setting the derivative of the function equal to zero and solving for $ x $.
2. Determine the nature of each critical point by analyzing the sign of the second derivative at each critical point.
3. Identify the relative maximum and minimum based on the nature of the critical points.

Let's go through these steps:

1. Find the derivative of $ f(x) $:
$$ f'(x) = 3x^2 - 6x - 1 $$

Set $ f'(x) $ equal to zero and solve for $ x $ to find the critical points:
$$ 3x^2 - 6x - 1 = 0 $$

Use the quadratic formula to solve for $ x $:
$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Plugging in the values $ a = 3 $, $ b = -6 $, and $ c = -1 $, we get:
$$ x = \frac{{6 \pm \sqrt{{(-6)^2 - 4(3)(-1)}}}}{{2(3)}} $$
$$ x = \frac{{6 \pm \sqrt{{36 + 12}}}}{{6}} $$
$$ x = \frac{{6 \pm \sqrt{{48}}}}{{6}} $$
$$ x = \frac{{6 \pm 4\sqrt{{3}}}}{{6}} $$

So, the critical points are:
$$ x_1 = \frac{{6 + 4\sqrt{{3}}}}{{6}} $$
$$ x_2 = \frac{{6 - 4\sqrt{{3}}}}{{6}} $$

2. Find the second derivative of $ f(x) $:
$$ f''(x) = 6x - 6 $$

3. Evaluate $ f''(x) $ at each critical point:
$$ f''\left(\frac{{6 + 4\sqrt{{3}}}}{{6}}\right) = 6\left(\frac{{6 + 4\sqrt{{3}}}}{{6}}\right) - 6 $$
$$ f''\left(\frac{{6 - 4\sqrt{{3}}}}{{6}}\right) = 6\left(\frac{{6 - 4\sqrt{{3}}}}{{6}}\right) - 6 $$

Determine the sign of $ f''(x) $ at each critical point to identify the nature of the critical points:

- If $ f''(x) > 0 $, the function has a relative minimum at that point.
- If $ f''(x) < 0 $, the function has a relative maximum at that point.

Finally, plug the critical points into the original function $ f(x) $ to find the corresponding $ y $-values, which represent the values of the function at the relative maximum and minimum points.