Question

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

a. (2 points) find all critical points.
b. (3 points) classify them as local maxima, local minima, or neither. you must use
a sign chart for full credit. *

Explanation:

Step1: Find the derivative of \(f(x)\)

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{3}\) and \(v=(1 - x)^{4}\).
\(u^\prime=3x^{2}\), \(v^\prime=-4(1 - x)^{3}\)
\(f^\prime(x)=3x^{2}(1 - x)^{4}-4x^{3}(1 - x)^{3}\)
Factor out \(x^{2}(1 - x)^{3}\):
\(f^\prime(x)=x^{2}(1 - x)^{3}[3(1 - x)-4x]=x^{2}(1 - x)^{3}(3 - 3x - 4x)=x^{2}(1 - x)^{3}(3 - 7x)\)

Step2: Find the critical points

Set \(f^\prime(x)=0\)
\(x^{2}(1 - x)^{3}(3 - 7x)=0\)
\(x^{2}=0\Rightarrow x = 0\)
\((1 - x)^{3}=0\Rightarrow x = 1\)
\(3-7x=0\Rightarrow x=\frac{3}{7}\)

Step3: Create a sign chart

Choose test points in the intervals \((-\infty,0)\), \((0,\frac{3}{7})\), \((\frac{3}{7},1)\) and \((1,\infty)\)

• For \(x=-1\): \(f^\prime(-1)=(-1)^{2}(1+ 1)^{3}(3 + 7)=1\times8\times10>0\)
• For \(x=\frac{1}{7}\): \(f^\prime(\frac{1}{7})=(\frac{1}{7})^{2}(1-\frac{1}{7})^{3}(3 - 1)=(\frac{1}{49})(\frac{6}{7})^{3}\times2>0\)
• For \(x=\frac{1}{2}\): \(f^\prime(\frac{1}{2})=(\frac{1}{2})^{2}(1-\frac{1}{2})^{3}(3-\frac{7}{2})=\frac{1}{4}\times\frac{1}{8}\times(-\frac{1}{2})<0\)
• For \(x = 2\): \(f^\prime(2)=2^{2}(1 - 2)^{3}(3-14)=4\times(-1)\times(-11)>0\)

Answer:

a. The critical points are \(x = 0\), \(x=\frac{3}{7}\) and \(x = 1\)
b.

• At \(x = 0\): Since \(f^\prime(x)\) does not change sign around \(x = 0\), \(x = 0\) is neither a local maximum nor a local minimum.
• At \(x=\frac{3}{7}\): Since \(f^\prime(x)\) changes sign from positive to negative around \(x=\frac{3}{7}\), \(x=\frac{3}{7}\) is a local maximum.
• At \(x = 1\): Since \(f^\prime(x)\) changes sign from negative to positive around \(x = 1\), \(x = 1\) is a local minimum.