How do you find the exact relative maximum and minimum of the polynomial function of #f(x) = x^3 - 2x^2 - x +1#?

Question

Charles Arocha · Accepted Answer

Find and test the critical numbers, then find #f(c)# for them.

#f(x) = x^3 - 2x^2 - x +1# #f'(x)=3x^2-4x-1# which is never undefined and is #0# at #x = (4+-sqrt(16+12))/6 = (2+-sqrt7)/3# The critical numbers are #(2-sqrt7)/3# and #(2+sqrt7)/3#. The sign of #f'(x)# is positive a little left of #(2-sqrt7)/3# and negative a little to the right. Therefore, #f((2-sqrt7)/3)# is a relative maximum. #f((2-sqrt7)/3) = ((2-sqrt7)/3)^3 - 2((2-sqrt7)/3)^2 - ((2-sqrt7)/3) +1# # = 7/(3sqrt3)(2sqrt7-1)# The sign of #f'(x)# is negative a little left of #(2+sqrt7)/3# and positive to the right. Therefore, #f((2+sqrt7)/3)# is a relative minimum. #f((2+sqrt7)/3) = ((2+sqrt7)/3)^3 - 2((2+sqrt7)/3)^2 - ((2+sqrt7)/3) +1# # = -7/(3sqrt3)(2sqrt7+1)#

William Abdalla · Answer

undefined To find the exact relative maximum and minimum of the polynomial function $ f(x) = x^3 - 2x^2 - x +1 $, 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 where the function is increasing or decreasing using the first derivative test.
3. Use the second derivative test to classify the critical points as relative maximums, relative minimums, or points of inflection.

Let's find the critical points first:

$ f'(x) = 3x^2 - 4x - 1 $

Setting $ f'(x) $ equal to zero and solving for $ x $:

$ 3x^2 - 4x - 1 = 0 $

The solutions for $ x $ will give the critical points.

Next, determine the intervals of increase and decrease by analyzing the sign of $ f'(x) $ in these intervals.

Finally, use the second derivative test on the critical points to determine whether they correspond to relative maximums, relative minimums, or points of inflection.