How do you find the local maximum and minimum values of # f(x)=x^3 + 6x^2 + 12x -1# using both the First and Second Derivative Tests?

Question

Isabella Ackerman · Accepted Answer

There are no minimum or maximum, only a point of inflection at #(-2,-9)#

Let's calculate the first and second derivatives

#f(x)=x^3+6x^2+12x-1#

#f'(x)=3x^2+12x+12#

#=3(x^2+4x+4)#

#=3(x+2)(x+2)#

#=3(x+2)^2#

and

#f''(x)=6x+12#

The critical point is when #f'(x)=0#

That is,

#x+2=0#

#x=-2#

Let's build a chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##f'(x)##color(white)(aaaaaa)##+##color(white)(aaaaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##↗##color(white)(aaaaaa)##↗#

Let's look at #f''(x)#

#f''(x)=0# when #x+2=0#

That is when #x=-2#

There is a point of inflection at #(-2,-9)#

Let's do a chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-2##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##f''(x)##color(white)(aaaaaa)##-##color(white)(aaaaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##nn##color(white)(aaaaaa)##uu# graph{x^3+6x^2+12x-1 [-38.27, 26.68, -22.07, 10.37]}

Emma Aldridge · Answer

undefined To find the local maximum and minimum values of $ f(x) = x^3 + 6x^2 + 12x - 1 $ using the First and Second Derivative Tests: 1. First, find the critical points by setting the derivative of $ f(x) $ equal to zero and solving for $ x $. $ f'(x) = 3x^2 + 12x + 12 $ Set $ f'(x) = 0 $ and solve for $ x $: $ 3x^2 + 12x + 12 = 0 $ $ x^2 + 4x + 4 = 0 $ $ (x + 2)^2 = 0 $ $ x = -2 $ 2. The critical point is $ x = -2 $. 3. Next, find the second derivative of $ f(x) $: $ f''(x) = 6x + 12 $ 4. Evaluate the second derivative at the critical point: $ f''(-2) = 6(-2) + 12 = 0 $ 5. Since the second derivative test is inconclusive, we resort to the First Derivative Test. Evaluate $ f'(x) $ to determine the behavior of the function around the critical point: When $ x < -2 $, $ f'(x) > 0 $, indicating the function is increasing. When $ x > -2 $, $ f'(x) < 0 $, indicating the function is decreasing. 6. Therefore, $ x = -2 $ corresponds to a local minimum point. 7. To find the minimum value, substitute $ x = -2 $ into $ f(x) $: $ f(-2) = (-2)^3 + 6(-2)^2 + 12(-2) - 1 $ $ f(-2) = -8 + 24 - 24 - 1 $ $ f(-2) = -9 $ 8. So, the local minimum value of $ f(x) $ is $ -9 $ occurring at $ x = -2 $.