How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #y=x^4-2x^3#?

Question

Isabella Ackerman · Accepted Answer

intercept / stationary point at x= 0 is an inflexion point

stationary point at #x = 3/2# is a relative minimum

#y=x^4-2x^3 = x^3(x-2)# so # y = 0# at #x = 0, 2# #y' = 4x^3 - 6x^2 = x^2(4x-6)# so #y' = 0# at #x = 0, 3/2# #y'' = 12x^2 - 12x = 12x(x-1) # #y''(0) = 0# and #y''(3/2) = 9 [> 0]# so the intercept and stationary point at x= 0 is an inflexion point and the stationary point at #x = 3/2# is a relative minimum globally, the dominant term in the expression is #x^4# so #lim_{x to pm oo} y = + oo# you can see all of this in the plot graph{x^4 - 2x^3 [-10, 10, -5, 5]}

Emma Aldridge · Answer

undefined To determine where the function $ y = x^4 - 2x^3 $ is increasing or decreasing, and to find relative maxima and minima: 1. Find the first derivative of the function, $ y' $, using the power rule. $$ y' = 4x^3 - 6x^2 $$ 2. Set $ y' = 0 $ to find critical points. $$ 4x^3 - 6x^2 = 0 $$ Factor out $ 2x^2 $: $$ 2x^2(2x - 3) = 0 $$ So, $ x = 0 $ and $ x = \frac{3}{2} $ are critical points. 3. Determine the intervals between these critical points and test points within these intervals to determine the sign of $ y' $. For $ x < 0 $: Choose $ x = -1 $ (test point). $$ y'(-1) = 4(-1)^3 - 6(-1)^2 = -4 - 6 = -10 $$ Since $ y'(-1) < 0 $, the function is decreasing on this interval. For $ 0 < x < \frac{3}{2} $: Choose $ x = 1 $ (test point). $$ y'(1) = 4(1)^3 - 6(1)^2 = 4 - 6 = -2 $$ Since $ y'(1) < 0 $, the function is decreasing on this interval. For $ x > \frac{3}{2} $: Choose $ x = 2 $ (test point). $$ y'(2) = 4(2)^3 - 6(2)^2 = 32 - 24 = 8 $$ Since $ y'(2) > 0 $, the function is increasing on this interval. 4. To find relative maxima and minima, examine the behavior of the function around the critical points. At $ x = 0 $: Since the function changes from decreasing to increasing at this point, it is a relative minimum. At $ x = \frac{3}{2} $: Since the function changes from increasing to decreasing at this point, it is a relative maximum. Therefore, the function $ y = x^4 - 2x^3 $ has a relative minimum at $ x = 0 $ and a relative maximum at $ x = \frac{3}{2} $.