How do you find the exact relative maximum and minimum of the polynomial function of #f(x)=2x^3+3x^2-12x #?

Question

Charles Arocha · Accepted Answer

Relative maximum: #f(-2)=20#
Relative minimum: #f(1) = -9# Given #color(white)("XXX")f(x)=2x^3+3x^2-12x# Note #color(white)("XXX")#Relative minimums/maximums happen points where #f'(x)=0# #f'(x) = 6x^2+6x-12# #color(white)("XXX")=6(x^2+x-2)# #color(white)("XXX")=6(x+2)(x-1)# #f'(x)=0# for # x=-2# and # x=+1# so these are the locations of the relative minimum/maximum values. #f(-2) = -16+12+24=20# #f(1) = 2+3-12 = -9# Therefore #color(white)("XXX")#the relative maximum is #20# (at #x=-2#) #color(white)("XXX")#the relative minimum is #(-9)# ( at #x=1#)

Isabella Ackerman · Answer

undefined To find the exact relative maximum and minimum of the polynomial function f(x) = 2x^3 + 3x^2 - 12x, you would first find the critical points by setting the derivative equal to zero and solving for x. Next, evaluate the second derivative at these critical points to determine the nature of the extremum (maximum or minimum). If the second derivative is positive, it's a relative minimum; if it's negative, it's a relative maximum. If the second derivative is zero or undefined, the test is inconclusive, and you would need to use other methods like the first derivative test or analyze the function graphically.