4. (5 points) for the function ( f(x)=x^{3}(1 - x)^{4} ),\na. (2 points) find all critical points.\nb. (3…

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.