hw 16 - first derivative test section 3.2 / 9\nprevious problem problem list next problem\nhw 16 - first…

hw 16 - first derivative test section 3.2 / 9\nprevious problem problem list next problem\nhw 16 - first derivative test section 3.2: problem 9\n(1 point)\nconsider the function\n$f(x)=\\ln (2 + x^{2})$ \n1. $f$ is increasing on the intervals\n2. $f$ is decreasing on the intervals\n3. the relative maxima of $f$ occur at $x=$\n4. the relative minima of $f$ occur at $x=$\nnotes: in the last two, your answer should be a comma separated list of $x$ values or the word \none\.\nnote: you can earn partial credit on this problem.\npreview my answers submit answers

hw 16 - first derivative test section 3.2 / 9\nprevious problem problem list next problem\nhw 16 - first derivative test section 3.2: problem 9\n(1 point)\nconsider the function\n$f(x)=\\ln (2 + x^{2})$ \n1. $f$ is increasing on the intervals\n2. $f$ is decreasing on the intervals\n3. the relative maxima of $f$ occur at $x=$\n4. the relative minima of $f$ occur at $x=$\nnotes: in the last two, your answer should be a comma separated list of $x$ values or the word \none\.\nnote: you can earn partial credit on this problem.\npreview my answers submit answers

Answer

Explanation:

Step1: Find the first derivative

Use the chain rule. If (y = \ln(u)) and (u=2 + x^{2}), then (\frac{dy}{dx}=\frac{1}{u}\cdot\frac{du}{dx}). Since (\frac{du}{dx} = 2x), the first derivative (f^{\prime}(x)=\frac{2x}{2 + x^{2}}).

Step2: Find critical points

Set (f^{\prime}(x)=0). (\frac{2x}{2 + x^{2}} = 0). Since (2 + x^{2}>0) for all real (x), then (2x = 0) gives (x = 0).

Step3: Test intervals

  • For (x<0) (e.g., (x=-1)), (f^{\prime}(-1)=\frac{2\times(-1)}{2+(-1)^{2}}=\frac{-2}{3}<0).
  • For (x>0) (e.g., (x = 1)), (f^{\prime}(1)=\frac{2\times1}{2 + 1^{2}}=\frac{2}{3}>0).

Answer:

  1. (f) is increasing on the interval ((0,\infty))
  2. (f) is decreasing on the interval ((-\infty,0))
  3. The relative maxima of (f) occur at (x=\text{none})
  4. The relative minima of (f) occur at (x = 0)