find the intervals on which the graph of ( f ) is concave upward, the intervals on which the graph of ( f )…

find the intervals on which the graph of ( f ) is concave upward, the intervals on which the graph of ( f ) is concave downward, and the inflection points.\n( f(x)=-x^{6}+12 x^{5}-12 x + 3 )\nfor what interval(s) of ( x ) is the graph of ( f ) concave upward? select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. \n(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed)\nb. the graph is never concave upward.

find the intervals on which the graph of ( f ) is concave upward, the intervals on which the graph of ( f ) is concave downward, and the inflection points.\n( f(x)=-x^{6}+12 x^{5}-12 x + 3 )\nfor what interval(s) of ( x ) is the graph of ( f ) concave upward? select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. \n(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed)\nb. the graph is never concave upward.

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ((x^n)^\prime = nx^{n - 1}), for (f(x)=-x^{6}+12x^{5}-12x + 3), we have (f^\prime(x)=-6x^{5}+60x^{4}-12).

Step2: Find the second derivative

Differentiate (f^\prime(x)) again. (f^{\prime\prime}(x)=-30x^{4}+240x^{3}=-30x^{3}(x - 8)).

Step3: Find the critical points of (f^{\prime\prime}(x))

Set (f^{\prime\prime}(x) = 0). (-30x^{3}(x - 8)=0). Solving this equation gives (x = 0) and (x = 8).

Step4: Test the intervals

  • For (x<0), let (x=-1). Then (f^{\prime\prime}(-1)=-30(-1)^{3}(-1 - 8)=-30\times(-1)\times(-9)=-270<0).
  • For (0<x<8), let (x = 1). Then (f^{\prime\prime}(1)=-30\times1^{3}(1 - 8)=-30\times(-7)=210>0).
  • For (x>8), let (x = 9). Then (f^{\prime\prime}(9)=-30\times9^{3}(9 - 8)=-30\times729\times1=-21870<0).

Answer:

A. ((0,8))