find all unique critical points for\n\n$f(x)=x^{4}\\ln (x)$.\n\nif there are fewer than 4 critical points…

find all unique critical points for\n\n$f(x)=x^{4}\\ln (x)$.\n\nif there are fewer than 4 critical points, leave the remaining field(s) blank.\n\nif a critcial point has multiplicity greater than 1, only enter it once.\n\n$x_{1}=$ number $x_{2}=$ number\n\n$x_{3}=$ number $x_{4}=$ number\n\nsave & grade 4 attempts left save only 9 points available for this attempt\n(following attempts are worth: 8, 6, 3)\n\nsubmitted answer\nparkerd4@illinois.edu submitted at 2025 - 10 - 28 18:42:14 (cdt) 0%\n\n$x_{1}=0$ $x_{2}=9999$\n$x_{3}=9999$ $x_{4}=9999$\n\nall your answers are incorrect.

find all unique critical points for\n\n$f(x)=x^{4}\\ln (x)$.\n\nif there are fewer than 4 critical points, leave the remaining field(s) blank.\n\nif a critcial point has multiplicity greater than 1, only enter it once.\n\n$x_{1}=$ number $x_{2}=$ number\n\n$x_{3}=$ number $x_{4}=$ number\n\nsave & grade 4 attempts left save only 9 points available for this attempt\n(following attempts are worth: 8, 6, 3)\n\nsubmitted answer\nparkerd4@illinois.edu submitted at 2025 - 10 - 28 18:42:14 (cdt) 0%\n\n$x_{1}=0$ $x_{2}=9999$\n$x_{3}=9999$ $x_{4}=9999$\n\nall your answers are incorrect.

Answer

Explanation:

Step1: Find the derivative of (f(x))

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{4}) and (v=\ln(x)). (u^\prime=4x^{3}), (v^\prime=\frac{1}{x}). So (f^\prime(x)=4x^{3}\ln(x)+x^{4}\cdot\frac{1}{x}=x^{3}(4\ln(x) + 1)).

Step2: Set (f^\prime(x)=0)

Since (x^{3}(4\ln(x)+1)=0). We have two cases:

  • Case 1: (x^{3}=0), but (x = 0) is not in the domain of (y=\ln(x)) (domain of (f(x)) is (x>0)).
  • Case 2: (4\ln(x)+1 = 0). Solve (4\ln(x)+1 = 0) for (x). First, (4\ln(x)=- 1), then (\ln(x)=-\frac{1}{4}). Using the property (y = \ln(x)\Leftrightarrow x = e^{y}), we get (x=e^{-\frac{1}{4}}).

Answer:

(x_1 = e^{-\frac{1}{4}})