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. ( (0,8) )\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.\nfor what interval(s) of ( x ) is the graph of ( f ) concave downward? select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. ( (-infty, 0),(8, infty) )\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 downward.\ndetermine the ( x ) coordinates of any inflection points of the graph of ( f(x) ). select the correct choice below and, if necessary, fill in the answer box to complete your chosen\na. ( x= ) (type an exact answer. use a comma to separate answers as needed)\nb. there are no inflection points.
Answer
Explanation:
Step1: Find the second derivative
First, find the first derivative of (f(x)=-x^{6}+12x^{5}-12x + 3) using the power rule ((x^{n})^\prime=nx^{n - 1}). (f^\prime(x)=-6x^{5}+60x^{4}-12). Then find the second derivative: (f^{\prime\prime}(x)=-30x^{4}+240x^{3}=-30x^{3}(x - 8)).
Step2: Determine concavity
Set (f^{\prime\prime}(x)=0), so (-30x^{3}(x - 8)=0), which gives (x = 0) and (x = 8). Test intervals:
- For (x\lt0), let (x=-1), then (f^{\prime\prime}(-1)=-30(-1)^{3}(-1 - 8)=-30\times(-1)\times(-9)=-270\lt0). The function is concave - downward on ((-\infty,0)).
- For (0\lt x\lt8), let (x = 1), then (f^{\prime\prime}(1)=-30\times1^{3}(1 - 8)=-30\times(-7)=210\gt0). The function is concave - upward on ((0,8)).
- For (x\gt8), let (x = 9), then (f^{\prime\prime}(9)=-30\times9^{3}(9 - 8)=-30\times729\times1=-21870\lt0). The function is concave - downward on ((8,\infty)).
Step3: Find inflection points
Since the concavity changes at (x = 0) and (x = 8), these are the (x) - coordinates of the inflection points.
Answer:
For concave upward: ((0,8)) For concave downward: ((-\infty,0)\cup(8,\infty)) For inflection points: (x = 0,x = 8)