Find the maxima and minima of function:- #y=2x^3-9x^2-24x+15#?

Question

Ezra Adams · Accepted Answer

Minima: #(4,-97)#
Maxima: #(-1,28)# #y=2x^3-9x^2-24x+15# #dy/dx=6x^2-18x-24# Let us determine the coordinates of the maxima and minima, When #dy/dx=0#, #6x^2-18x-24=0# #x^2-3x-4=0# Factor and solve, #(x-4)(x+1)=0# #x=4 or -1# When #x=4#, #y=2(4)^3-9(4)^2-24(4)+15# #color(white)(y)=-97# #(4,-97)# When #x=-1# #y=2(-1)^3-9(-1)^2-24(-1)+15# #color(white)(y)=28# #(-1,28)# Now, to determine the nature of these coordinates, Find #(d^2y)/dx^2#, #(d^2y)/dx^2=12x-18# When #x=4#, #(d^2y)/dx^2=12(4)-18# #color(white)((d^2y)/dx^2)=30>0# ( minima ) When #x=-1#, #(d^2y)/dx^2=12(-1)-18# #color(white)((d^2y)/dx^2)=-30<0# ( maxima ) Therefore, #(4,-97)# minima and #(-1,28)# maxima Check: graph{2x^3-9x^2-24x+15 [-20, 20, -120,120]}

Christopher Allers · Answer

undefined To find the maxima and minima of the function y = 2x^3 - 9x^2 - 24x + 15, we first find the critical points by taking the derivative of the function and setting it equal to zero. Then, we determine whether these critical points correspond to maxima or minima by analyzing the sign changes of the derivative.

First, find the derivative of the function:
y' = 6x^2 - 18x - 24

Set y' equal to zero and solve for x to find the critical points:
6x^2 - 18x - 24 = 0
Factor out the common factor of 6:
6(x^2 - 3x - 4) = 0
(x^2 - 3x - 4) = 0
Factor the quadratic equation:
(x - 4)(x + 1) = 0
Solve for x:
x = 4 or x = -1

These are the critical points. Now, we will use the second derivative test to determine whether they correspond to maxima or minima.

Take the second derivative of the function:
y'' = 12x - 18

Now, substitute the critical points into the second derivative:
At x = 4: y''(4) = 12(4) - 18 = 48 - 18 = 30 (positive, so it's a local minimum)
At x = -1: y''(-1) = 12(-1) - 18 = -12 - 18 = -30 (negative, so it's a local maximum)

Therefore, the function has a local minimum at x = 4 and a local maximum at x = -1. To find the corresponding y-values, plug these x-values back into the original function:

At x = 4: y(4) = 2(4)^3 - 9(4)^2 - 24(4) + 15 = 2(64) - 9(16) - 96 + 15 = 128 - 144 - 96 + 15 = -97
At x = -1: y(-1) = 2(-1)^3 - 9(-1)^2 - 24(-1) + 15 = 2(-1) - 9(1) + 24 + 15 = -2 - 9 + 24 + 15 = 28

So, the local minimum is (-1, 28) and the local maximum is (4, -97).