How do you find the intervals of increasing and decreasing using the first derivative given #y=(x+2)^2(x-1)#?

Question

Christopher Agresta · Accepted Answer

Using the product rule:

#dy/dx = 2(x+2)(x-1) +(x+2)^2 = (x+2)(2x-2+x+2) = 3x(x+2)#

So:

#dy/dx < 0 # for #x in (-2,0)#

#dy/dx > 0 # for #in in (-oo,-2) uu (0,+oo)#

which means that #y(x)# is increasing from #-oo# to #x=-2# where it has a local maximum, decreasing from #x=-2# to #x=0# where it has a local minimum, and again increasing from #x=0# to #+oo#.

graph{(x+2)^2(x-1) [-4, 3, -10, 10]}

Axel Alvarez · Answer

undefined To find the intervals of increasing and decreasing for the function y = (x + 2)^2(x - 1), you first need to find its first derivative. 1. Find the first derivative of y with respect to x: y' = d/dx [(x + 2)^2(x - 1)] 2. Apply the product rule and chain rule to differentiate the function: y' = 2(x + 2)(x - 1) + (x + 2)^2 3. Simplify the expression: y' = 2(x^2 - x + 2x - 2) + (x^2 + 4x + 4) = 2x^2 - 2x + 4x - 4 + x^2 + 4x + 4 = 3x^2 + 6x Now, to find the intervals of increasing and decreasing: - Set y' = 0 to find critical points. - Determine the sign of the derivative in the intervals between critical points. 4. Set y' = 0 and solve for x: 3x^2 + 6x = 0 x(3x + 6) = 0 x = 0 or x = -2 5. Determine the intervals: - Pick test points from each interval to determine the sign of the derivative. - Test a value less than -2, between -2 and 0, and greater than 0 in y'. 6. Test point less than -2: Choose x = -3. y' = 3(-3)^2 + 6(-3) = 27 - 18 = 9, which is positive. 7. Test point between -2 and 0: Choose x = -1. y' = 3(-1)^2 + 6(-1) = 3 - 6 = -3, which is negative. 8. Test point greater than 0: Choose x = 1. y' = 3(1)^2 + 6(1) = 3 + 6 = 9, which is positive. Based on the signs of the derivative: - y' > 0: Function is increasing. - y' < 0: Function is decreasing. So, the intervals are: - Increasing: x < -2 and x > 0 - Decreasing: -2 < x < 0