5. (5 points) for the function ( f(x)=3 x^{4}-4 x^{3}+6 ),\na. (2 points) find all the candidates for…

Explanation:

Step1: Find the second derivative

First, find the first derivative (f^\prime(x)) using the power rule ((x^n)^\prime = nx^{n - 1}). (f^\prime(x)=\frac{d}{dx}(3x^{4}-4x^{3}+6)=12x^{3}-12x^{2}) Then, find the second derivative (f^{\prime\prime}(x)) by differentiating (f^\prime(x)) again. (f^{\prime\prime}(x)=\frac{d}{dx}(12x^{3}-12x^{2}) = 36x^{2}-24x=12x(3x - 2))

Step2: Find candidates for inflection points

Set (f^{\prime\prime}(x)=0). (12x(3x - 2)=0) Using the zero - product property (ab = 0\Rightarrow a = 0) or (b = 0), we have (x = 0) or (x=\frac{2}{3})

Step3: Create a sign chart for (f^{\prime\prime}(x))

• Choose test intervals: ((-\infty,0)), ((0,\frac{2}{3})), ((\frac{2}{3},\infty))
• For (x=-1) (in ((-\infty,0))): (f^{\prime\prime}(-1)=12\times(-1)\times(3\times(-1)-2)=12\times(-1)\times(-5) = 60>0)
• For (x=\frac{1}{3}) (in ((0,\frac{2}{3}))): (f^{\prime\prime}(\frac{1}{3})=12\times\frac{1}{3}\times(3\times\frac{1}{3}-2)=4\times(-1)=-4<0)
• For (x = 1) (in ((\frac{2}{3},\infty))): (f^{\prime\prime}(1)=12\times1\times(3\times1 - 2)=12\times1\times1=12>0)

Answer:

a. The candidates for inflection points are (x = 0) and (x=\frac{2}{3}) b. Since the concavity of (y = f(x)) changes at (x = 0) (from concave up ((f^{\prime\prime}(x)>0) for (x<0)) to concave down ((f^{\prime\prime}(x)<0) for (0<x<\frac{2}{3}))) and at (x=\frac{2}{3}) (from concave down ((f^{\prime\prime}(x)<0) for (0<x<\frac{2}{3})) to concave up ((f^{\prime\prime}(x)>0) for (x>\frac{2}{3}))), both (x = 0) and (x=\frac{2}{3}) are inflection points.