Question

identify the coordinates of any local and absolute extreme points and inflection points. graph the function.

$y = 12x^{3}-x^{4}=x^{3}(12 - x)$

(use a comma to separate answers as needed. type an ordered pair. do not use commas in the individual coordinates.)

b. there are no inflection points.

graph the function. choose the correct answer below.

a.
b.
c.
d.

Explanation:

Step1: Find the first derivative

Using the power rule \((x^n)^\prime = nx^{n - 1}\), for \(y = 12x^3-x^4\), \(y^\prime=36x^2 - 4x^3=4x^2(9 - x)\)
Set \(y^\prime = 0\), \(4x^2(9 - x)=0\). Solving \(4x^2=0\) gives \(x = 0\), and solving \(9 - x=0\) gives \(x = 9\)

Step2: Find the second derivative

Differentiate \(y^\prime=36x^2 - 4x^3\) using the power rule. \(y^{\prime\prime}=72x-12x^2 = 12x(6 - x)\)
Set \(y^{\prime\prime}=0\), \(12x(6 - x)=0\). Solving \(12x=0\) gives \(x = 0\), and solving \(6 - x=0\) gives \(x = 6\)

Step3: Analyze critical points

• For \(x = 0\):

\(y(0)=12\times0^3-0^4 = 0\). The second - derivative test: \(y^{\prime\prime}(0)=0\). Using the first - derivative test, \(y^\prime\) does not change sign around \(x = 0\) (since \(y^\prime=4x^2(9 - x)\) and \(x^2\geq0\) for all \(x\)), so \(x = 0\) is not a local extremum.

• For \(x = 9\):

\(y(9)=12\times9^3-9^4=12\times729 - 6561=8748-6561 = 2187\).
Since \(y^{\prime\prime}(9)=72\times9-12\times9^2=648 - 972=- 324<0\), \(x = 9\) is a local maximum.

• For \(x = 6\):

\(y(6)=12\times6^3-6^4=12\times216-1296=2592 - 1296 = 1296\). Since \(y^{\prime\prime}\) changes sign around \(x = 6\) ( \(y^{\prime\prime}(x)\) is positive for \(x\in(0,6)\) and negative for \(x\in(6,\infty)\) when \(x>0\)), \((6,1296)\) is an inflection point.

Answer:

Local maximum at \((9,2187)\), inflection point at \((6,1296)\)