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

Question

Charles Anctil · Accepted Answer

#(-2,18)# is relative maximum point and
#(2,-14)# is relative minimum point.

#f(x)=x^3-12 x +2# #f^'(x)=3 x^2-12 #, critical points are those point where, slope,#f'(x)=0 :. 3 x^2-12=0 or 3 (x^2-4)=0 #or #3 (x+2)(x-2)=0 or x= -2 and x=2# #f(-2)= (-2)^3-12*(-2)+2= 18 or (-2,18)# and #f(2)= 2^3-12*2+2= -14 or (2, -14)# Slope check in interval , # x< -2 , f^'(x)= (-)*(-)=(+) ; f'(x)>0 :. # Therefore, increasing slope. # -2< x<2 , f^'(x)= (+)*(-)=(-) ; f'(x)<0 :. #, Therefore, decreasing slope . # x> 2 , f^'(x)= (+)*(+)=(+) ; f'(x)>0 :. # Therefore, increasing slope. At #x=-2# the slope changes from increasing to decreasing, so #(-2,18)# is local maximum point and at #x=2# the slope changes from decreasing to increasing, hence, #(2,-14)# is local minimum point. x^3–12x +2 [-40, 40, -20, 20]} [Ans]

Charles Arocha · Answer

undefined To find the exact relative maximum and minimum of the polynomial function $ f(x) = x^3 - 12x + 2 $, follow these steps:

1. Find the critical points by taking the derivative of the function and setting it equal to zero.
2. Determine the intervals of increase and decrease using the first derivative test.
3. Identify the points of inflection by finding the second derivative.
4. Use the second derivative test to classify the critical points as relative maxima, minima, or points of inflection.