How do you use the first and second derivatives to sketch #y = x^3 - 12x - 12#?

Question

Aurora Alvarado · Accepted Answer

See the answer bellow.

#y=x^3-12x-12# First derivative: #y'=3x^2-12# Second derivative: #y''=6x# Using #y'=3x^2-12# #3(x^2-4)# #3(x-2)(x+2)# #x=-2,2# Sub x values into #y=x^3-12x-12#, to find y values #y=(-2)^3-12(-2)-12# #y=4#, #(-2,4)# #y=(2)^3-12(2)-12# #y=-28#, #(2,-28)# Using #y''=6x# Sub x values into #y''=6x# #6(-2)=-12# #x<0# #:.# #(2,-28)# is a maximum point #6(2)=12# #x>0# #:.# #(-2,4)# is a minimum point When #x=0# #y=0^3-12(0)-12# #y=-12# #(0,-12)# Use the information from the first and second derivative to sketch the graph as well as the y intercept. graph{y=x^3-12x-12 [-10, 10, -5, 5]}

Emma Aldridge · Answer

undefined To sketch the graph of $ y = x^3 - 12x - 12 $ using the first and second derivatives, follow these steps:

1. Find the first derivative $ y' $ by differentiating the function $ y $ with respect to $ x $.
$$ y' = 3x^2 - 12 $$

2. Set $ y' = 0 $ and solve for $ x $ to find critical points.
$$ 3x^2 - 12 = 0 $$
$$ x^2 = 4 $$
$$ x = \pm 2 $$

3. Test the intervals between and around the critical points using the first derivative test to determine the increasing and decreasing intervals.

4. Find the second derivative $ y'' $ by differentiating $ y' $ with respect to $ x $.
$$ y'' = 6x $$

5. Determine the concavity of the graph by analyzing the sign of the second derivative in each interval.

6. Use the critical points, increasing/decreasing intervals, and concavity information to sketch the graph of the function $ y = x^3 - 12x - 12 $.