Question

let ( f(x)=x^{4}(x - 4)^{3} ).
(a) find the critical numbers of the function ( f ). (enter your answers from smallest to largest.)
smallest value ( x_{1}=)
( x_{2}=)
largest value ( x_{3}=)
(b) what does the second derivative test tell you about the behavior of ( f ) at these critical numbers?
at ( x_{1} ) the second derivative test
at ( x_{2} ) the second derivative test
at ( x_{3} ) the second derivative test
(c) what does the first derivative test tell you? note what the first derivative test tells you that second derivative tes
at ( x_{1} ) the first derivative test
at ( x_{2} ) the first derivative test
at ( x_{3} ) the first derivative test

Explanation:

Step1: Find the first derivative

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{4}\) and \(v=(x - 4)^{3}\).
\(u^\prime=4x^{3}\), \(v^\prime = 3(x - 4)^{2}\)
\(f^\prime(x)=4x^{3}(x - 4)^{3}+x^{4}\times3(x - 4)^{2}\)
Factor out \(x^{3}(x - 4)^{2}\):
\(f^\prime(x)=x^{3}(x - 4)^{2}[4(x - 4)+3x]\)
Simplify the expression inside the brackets:
\(4(x - 4)+3x=4x-16 + 3x=7x-16\)
So \(f^\prime(x)=x^{3}(x - 4)^{2}(7x - 16)\)

Step2: Find the critical numbers

Set \(f^\prime(x)=0\)
\(x^{3}(x - 4)^{2}(7x - 16)=0\)
\(x^{3}=0\Rightarrow x = 0\)
\((x - 4)^{2}=0\Rightarrow x = 4\)
\(7x-16=0\Rightarrow x=\frac{16}{7}\approx2.29\)

Answer:

\(x_{1}=0\), \(x_{2}=\frac{16}{7}\), \(x_{3}=4\)