Question

1. (5 points) for the function ( f(x)=3 x^{4}-4 x^{3}+6 ),

a. (2 points) find all the candidates for inflection points.
b. (3 points) classify them as an inflection point or not. you must use a sign chart for full credit.

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 \(0\frac{2}{3}\))), both \(x = 0\) and \(x=\frac{2}{3}\) are inflection points.