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

Question

Aurora Alvarado · Accepted Answer

Maximum #=(-1,7)#
Minimum #=(2,-20)# Since the graph is not limited by a definition, we can just look for the extremal points through the dervivate of the function. #2x^3-3x^2-12x# Dervivating the function with rule #(x^n)'=nx^(n-1)#, will give. #(2x^3-3x^2-12x)'=6x^2-6x-12# After finding the #f'(x)# we can look for zero points of that graph. Because this will be equal to the extremal points. #6x^2-6x-12=0# From here we can take several paths. Such as either use the quadratic equation or factorize to find the zero points. This will give us. #x=2# and #x=-1# which can be seen as #(x-2)(x+1)#. We can now put these numbers into the original function #f(x)# and find the #y-#values of the extremal points. #f(2)=2*2^3-3*2^2-12*2=-20# #f(-1)=2*(-1)^3-3*(-1)^2-12*(-1)=7# These will be equal to coordinates, and we can with logic say which is which by this calculation. Since #f(2)=-20# we know that this has to be equal to the minimum point. And that #f(-1)=7# has to be equal to the maximum point. So by that, we can say that. Maximum #=(-1,7)# Minimum #=(2,-20)#

Samantha Adkison · Answer

undefined To find the exact relative maximum and minimum of the polynomial function $ f(x) = 2x^3 - 3x^2 - 12x $, we first need to find the critical points by taking the derivative of the function and setting it equal to zero.

1. Find the derivative of $ f(x) $:
   $ f'(x) = 6x^2 - 6x - 12 $.

2. Set the derivative equal to zero and solve for $ x $ to find the critical points:
   $ 6x^2 - 6x - 12 = 0 $.
   Simplifying, we get $ x^2 - x - 2 = 0 $.
   Factoring, we get $ (x + 1)(x - 2) = 0 $.
   So, the critical points are $ x = -1 $ and $ x = 2 $.

3. To determine if these critical points are relative maxima or minima, we use the second derivative test.

a. Find the second derivative of $ f(x) $:
      $ f''(x) = 12x - 6 $.

b. Evaluate $ f''(-1) $ and $ f''(2) $ to determine the concavity at the critical points.
      $ f''(-1) = 12(-1) - 6 = -18 $ (negative, so concave down at $ x = -1 $, potential relative maximum).
      $ f''(2) = 12(2) - 6 = 18 $ (positive, so concave up at $ x = 2 $, potential relative minimum).

4. Therefore, the function has a relative maximum at $ x = -1 $ and a relative minimum at $ x = 2 $.