Question

first derivative test: problem 3
(1 point)
the function ( f(x)=-2 x^{3}+33 x^{2}-180 x + 2 ) has one local minimum and one local maximum.
this function has a local minimum at ( x ) equals with value
and a local maximum at ( x ) equals with value
note: you can earn partial credit on this problem.
preview my answers submit answers
you have attempted this problem 0 times.
you have unlimited attempts remaining.
email instructor
page generated october 22, 2025 at 2:36:13 pm cdt
webwork © 1996 - 2024 | theme math4 | ww_version 2.19 | pg_version 2.19
the webwork project

Explanation:

Step1: Find the first derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(f(x)=-2x^{3}+33x^{2}-180x + 2\), we have \(f^\prime(x)=-6x^{2}+66x - 180\).
Factor out \(-6\): \(f^\prime(x)=-6(x^{2}-11x + 30)\).
Factor the quadratic: \(f^\prime(x)=-6(x - 5)(x - 6)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\). Then \(-6(x - 5)(x - 6)=0\).
Solving \(x-5 = 0\) gives \(x = 5\), and solving \(x - 6=0\) gives \(x = 6\).

Step3: Use the first - derivative test

Choose test points:

• For \(x\lt5\), let \(x = 4\). Then \(f^\prime(4)=-6(4 - 5)(4 - 6)=-6\times(-1)\times(-2)=-12\lt0\).
• For \(5\lt x\lt6\), let \(x = 5.5\). Then \(f^\prime(5.5)=-6(5.5 - 5)(5.5 - 6)=-6\times0.5\times(-0.5)=1.5\gt0\).
• For \(x\gt6\), let \(x = 7\). Then \(f^\prime(7)=-6(7 - 5)(7 - 6)=-6\times2\times1=-12\lt0\).

Since \(f^\prime(x)\) changes sign from negative to positive at \(x = 5\), \(x = 5\) is a local minimum.
Since \(f^\prime(x)\) changes sign from positive to negative at \(x = 6\), \(x = 6\) is a local maximum.

Step4: Find the function values

For \(x = 5\):
\(f(5)=-2\times5^{3}+33\times5^{2}-180\times5 + 2=-2\times125+33\times25-900 + 2=-250+825-900 + 2=-323\).
For \(x = 6\):
\(f(6)=-2\times6^{3}+33\times6^{2}-180\times6 + 2=-2\times216+33\times36-1080 + 2=-432+1188-1080 + 2=-322\).

Answer:

The function has a local minimum at \(x = 5\) with value \(-323\) and a local maximum at \(x = 6\) with value \(-322\).