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hw 16 - first derivative test section 3.2 / 9
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hw 16 - first derivative test section 3.2: problem 9
(1 point)
consider the function
$f(x)=\ln (2 + x^{2})$
- $f$ is increasing on the intervals
- $f$ is decreasing on the intervals
- the relative maxima of $f$ occur at $x=$
- the relative minima of $f$ occur at $x=$
notes: in the last two, your answer should be a comma separated list of $x$ values or the word
one\.
note: you can earn partial credit on this problem.
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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., \(x = 1\)), \(f^{\prime}(1)=\frac{2\times1}{2 + 1^{2}}=\frac{2}{3}>0\).
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- \(f\) is increasing on the interval \((0,\infty)\)
- \(f\) is decreasing on the interval \((-\infty,0)\)
- The relative maxima of \(f\) occur at \(x=\text{none}\)
- The relative minima of \(f\) occur at \(x = 0\)