first derivative test: problem 3\n(1 point)\nthe function ( f(x)=-2 x^{3}+33 x^{2}-180 x + 2 ) has one local…

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

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

Answer

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).