Question

let (f(x)=(x - 3)^{3}+4). use a graphing calculator (like desmos) to graph the function (f).
a. determine the interval(s) of the domain over which (f) has positive concavity (or the graph is \concave up\)
3,oo
preview
((3,infty)=(3,infty))
b. determine the interval(s) of the domain over which (f) has negative concavity (or the graph is \concave down\)

c. determine any inflection points for the function. format your answer(s) as an ordered pair. if there are multiple inflection points, enter all of them as a comma - separated list.

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Explanation:

Step1: Find the second - derivative of the function

First, expand \(f(x)=(x - 3)^{3}+4=x^{3}-9x^{2}+27x - 27 + 4=x^{3}-9x^{2}+27x-23\).
The first - derivative \(f^\prime(x)=3x^{2}-18x + 27\).
The second - derivative \(f^{\prime\prime}(x)=6x-18\).

Step2: Find where \(f^{\prime\prime}(x)>0\) for positive concavity

Set \(f^{\prime\prime}(x)>0\), so \(6x - 18>0\).
Add 18 to both sides: \(6x>18\).
Divide both sides by 6: \(x > 3\). So the interval of positive concavity is \((3,\infty)\).

Step3: Find where \(f^{\prime\prime}(x)<0\) for negative concavity

Set \(f^{\prime\prime}(x)<0\), so \(6x - 18<0\).
Add 18 to both sides: \(6x<18\).
Divide both sides by 6: \(x<3\). So the interval of negative concavity is \((-\infty,3)\).

Step4: Find inflection points

Set \(f^{\prime\prime}(x) = 0\), so \(6x-18 = 0\).
Add 18 to both sides: \(6x=18\).
Divide both sides by 6: \(x = 3\).
When \(x = 3\), \(f(3)=(3 - 3)^{3}+4=4\). So the inflection point is \((3,4)\).

Answer:

a. \((3,\infty)\)
b. \((-\infty,3)\)
c. \((3,4)\)