let ( f(x)=x^{4}(x - 4)^{3} ).\n(a) find the critical numbers of the function ( f ). (enter your answers…

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

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

Answer

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)