How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #f(x) = 3x^4 + 16x^3 + 24x^2 + 32#?

Question

Emilia Aguilar · Accepted Answer

See the answer below:

Credits:
Thanks to the web site wich helped us to remind the subject of increase and decrease of a function.

Aurora Alvarado · Answer

undefined To determine where the function is increasing or decreasing and to find the relative maxima and minima for $ f(x) = 3x^4 + 16x^3 + 24x^2 + 32 $, follow these steps:

1. Find the first derivative of the function $ f'(x) $.
2. Set $ f'(x) $ equal to zero to find critical points.
3. Determine the sign of $ f'(x) $ in the intervals defined by the critical points to identify where the function is increasing or decreasing.
4. Use the second derivative test or examine the behavior of $ f'(x) $ around critical points to determine the nature of relative extrema (maxima or minima).

Let's go through each step:

1. Find the first derivative:
$$ f'(x) = 12x^3 + 48x^2 + 48x $$

2. Set $ f'(x) $ equal to zero and solve for $ x $ to find critical points:
$$ 12x^3 + 48x^2 + 48x = 0 $$
$$ 12x(x^2 + 4x + 4) = 0 $$
$$ 12x(x + 2)^2 = 0 $$

This equation has one critical point at $ x = -2 $.

3. Determine the sign of $ f'(x) $ in the intervals defined by the critical point and endpoints of the domain (if any). This helps identify where the function is increasing or decreasing.

Consider the intervals $ (-\infty, -2) $, $ (-2, \infty) $:

For $ (-\infty, -2) $:
- Test a value less than -2, say -3, into $ f'(x) $:
  $ f'(-3) = 12(-3)(-3+2)^2 = 12(-3)(1)^2 = -36 < 0 $
  So, $ f'(x) $ is negative in this interval, indicating the function is decreasing.

For $ (-2, \infty) $:
- Test a value greater than -2, say 0, into $ f'(x) $:
  $ f'(0) = 12(0)(0+2)^2 = 0 $
  So, $ f'(x) $ is zero at $ x = -2 $, and $ f'(x) $ is positive for $ x > -2 $, indicating the function is increasing.

4. Use the second derivative test or examine the behavior of $ f'(x) $ around critical points to determine the nature of relative extrema.

The second derivative of $ f(x) $ is:
$$ f''(x) = 36x^2 + 96x + 48 $$

Evaluate $ f''(-2) $:
$$ f''(-2) = 36(-2)^2 + 96(-2) + 48 = 144 - 192 + 48 = 0 $$

Since $ f''(-2) = 0 $ and $ f''(x) $ changes sign from positive to negative, there is a relative maximum at $ x = -2 $.

Therefore, the function $ f(x) $ is decreasing on $ (-\infty, -2) $, has a relative maximum at $ x = -2 $, and is increasing on $ (-2, \infty) $.