How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #f (x) = x^3 + 6x^2#?

Question

How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #f (x)  =  x^3 + 6x^2#?

Cameron Alexander · Accepted Answer

Find when #f'(x)=0# or #"DNE"#. #f'(x)=3x^2+12x=3x(x+4)# #f'(x)=0# when #x=-4,0#.
#f'(x)# never #"DNE"#. Now, use a sign chart with #-4,0#. #f'(x)color(white)(xxxxxxxx)-4color(white)(xxxxxxxxxxxxx)0#
#larr------------------rarr# 
#color(white)(xxxx)"POSITIVE"color(white)(xxxxx)"NEGATIVE"color(white)(xxxxxx)"POSITIVE"# #f# is increasing whenever #f'(x)>0#.
#f# is decreasing whenever #f'(x)<0#. Thus, #f# is increasing on #(-oo,-4)uu(0,+oo)#.
#f# is decreasing on #(-4,0)#. A relative maximum occurs whenever #f'# switches from positive to negative.
A relative minimum occurs whenever #f'# switches from negative to positive. Thus, There is a relative maximum when #x=-4#.
There is a relative minimum when #x=0#. graph{x^3+6x^2 [-51.76, 65.27, -14.2, 44.35]}

Axel Alvarez · Answer

undefined To determine where the function is increasing or decreasing and to find the relative maxima and minima of $ f(x) = x^3 + 6x^2 $, follow these steps:

1. Find the derivative of the function, $ f'(x) $.
2. Set $ f'(x) $ equal to zero to find critical points.
3. Use the first derivative test to determine where the function is increasing or decreasing.
4. Use the second derivative test or analyze the behavior around critical points to determine relative maxima and minima.

Let's solve it step by step:

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

2. Set $ f'(x) $ equal to zero and solve for critical points:
   $ 3x^2 + 12x = 0 $.
   Factor out $ 3x $: $ 3x(x + 4) = 0 $.
   So, $ x = 0 $ and $ x = -4 $.

3. Use the first derivative test to determine where the function is increasing or decreasing:
   Plug test points into $ f'(x) $ in the intervals $ (-\infty, -4) $, $ (-4, 0) $, and $ (0, \infty) $.
   For example, test points could be $ x = -5 $, $ x = -2 $, and $ x = 1 $.
   $ f'(-5) = 3(-5)^2 + 12(-5) = 75 - 60 = 15 $, so $ f'(x) > 0 $ on $ (-\infty, -4) $, indicating the function is increasing on this interval.
   $ f'(-2) = 3(-2)^2 + 12(-2) = 12 - 24 = -12 $, so $ f'(x) < 0 $ on $ (-4, 0) $, indicating the function is decreasing on this interval.
   $ f'(1) = 3(1)^2 + 12(1) = 3 + 12 = 15 $, so $ f'(x) > 0 $ on $ (0, \infty) $, indicating the function is increasing on this interval.

4. Use the second derivative test or analyze the behavior around critical points to determine relative maxima and minima:
   Evaluate the second derivative, $ f''(x) = 6x + 12 $.
   At $ x = -4 $, $ f''(-4) = 6(-4) + 12 = -24 + 12 = -12 $, which is less than zero, indicating a relative maximum at $ x = -4 $.
   At $ x = 0 $, $ f''(0) = 6(0) + 12 = 12 $, which is greater than zero, indicating a relative minimum at $ x = 0 $.

So, the function $ f(x) = x^3 + 6x^2 $ is increasing on $ (-\infty, -4) $ and $ (0, \infty) $ and decreasing on $ (-4, 0) $. It has a relative maximum at $ x = -4 $ and a relative minimum at $ x = 0 $.