this is similar to section 3.4 problem 26: qianbuscalc2 3.4.026.r for the function s(x)=4x^3 - x^4, do the…

this is similar to section 3.4 problem 26: qianbuscalc2 3.4.026.r for the function s(x)=4x^3 - x^4, do the following: (a) determine its critical number(s) if exists. list from least value. answer: x_c1= x_c2= (b) use the first - derivative test to classify each as a relative maximum or minimum, or neither: near the critical number x_c1=, when x<x_c1, the first derivative is ---select---, and when x>x_c1, the first derivative is ---select---. therefore the function has ---select--- at x_c1=. near the critical number x_c2=, when x<x_c2, the first derivative is ---select---, and when x>x_c2, the first derivative is ---select---. therefore the function has ---select--- at x_c2=. hint: follow example 4. symbolic formatting help resources ebook_

this is similar to section 3.4 problem 26: qianbuscalc2 3.4.026.r for the function s(x)=4x^3 - x^4, do the following: (a) determine its critical number(s) if exists. list from least value. answer: x_c1= x_c2= (b) use the first - derivative test to classify each as a relative maximum or minimum, or neither: near the critical number x_c1=, when x<x_c1, the first derivative is ---select---, and when x>x_c1, the first derivative is ---select---. therefore the function has ---select--- at x_c1=. near the critical number x_c2=, when x<x_c2, the first derivative is ---select---, and when x>x_c2, the first derivative is ---select---. therefore the function has ---select--- at x_c2=. hint: follow example 4. symbolic formatting help resources ebook_

Answer

Explanation:

Step1: Find the derivative of the function

The derivative of $s(x)=4x^{3}-x^{4}$ using the power - rule $(x^n)' = nx^{n - 1}$ is $s'(x)=12x^{2}-4x^{3}=4x^{2}(3 - x)$.

Step2: Find the critical numbers

Set $s'(x)=0$. Then $4x^{2}(3 - x)=0$. Using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So $4x^{2}=0$ gives $x = 0$ and $3 - x=0$ gives $x = 3$. The critical numbers are $x_{c1}=0$ and $x_{c2}=3$.

Step3: Apply the First - Derivative Test for $x_{c1}=0$

Choose a test point to the left of $x = 0$, say $x=-1$. Then $s'(-1)=4\times(-1)^{2}\times(3+1)=16>0$. Choose a test point to the right of $x = 0$, say $x = 1$. Then $s'(1)=4\times1^{2}\times(3 - 1)=8>0$. Since the sign of the first - derivative does not change at $x = 0$, the function has neither a relative maximum nor a relative minimum at $x = 0$.

Step4: Apply the First - Derivative Test for $x_{c2}=3$

Choose a test point to the left of $x = 3$, say $x = 2$. Then $s'(2)=4\times2^{2}\times(3 - 2)=16>0$. Choose a test point to the right of $x = 3$, say $x = 4$. Then $s'(4)=4\times4^{2}\times(3 - 4)=-64<0$. Since the first - derivative changes sign from positive to negative at $x = 3$, the function has a relative maximum at $x = 3$.

Answer:

(a) $x_{c1}=0$, $x_{c2}=3$ (b) Near the critical number $x_{c1}=0$, when $x < x_{c1}$, the first derivative is positive, and when $x>x_{c1}$, the first derivative is positive. Therefore the function has neither at $x_{c1}=0$. Near the critical number $x_{c2}=3$, when $x < x_{c2}$, the first derivative is positive, and when $x>x_{c2}$, the first derivative is negative. Therefore the function has a relative maximum at $x_{c2}=3$.