Question

hw 16 - first derivative test section 3.2: problem 5
(1 point)
consider the function
$f(x)=x^{3}-9x^{2}+15x + 3$.

1. $f$ is increasing on the intervals
2. $f$ is decreasing on the intervals
3. the relative maxima of $f$ occur at $x=$
4. 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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Explanation:

Step1: Find the first derivative

The derivative of \(f(x)=x^{3}-9x^{2}+15x + 3\) is \(f^\prime(x)=3x^{2}-18x + 15\). Factor \(f^\prime(x)\): \(f^\prime(x)=3(x^{2}-6x + 5)=3(x - 1)(x - 5)\).

Step2: Find critical points

Set \(f^\prime(x)=0\), then \(3(x - 1)(x - 5)=0\). Solving for \(x\), we get \(x = 1\) and \(x = 5\).

Step3: Determine intervals of increase and decrease

Use test - points.

• For \(x\lt1\), let \(x = 0\). Then \(f^\prime(0)=3(0 - 1)(0 - 5)=15\gt0\). So \(f(x)\) is increasing on \((-\infty,1)\).
• For \(1\lt x\lt5\), let \(x = 2\). Then \(f^\prime(2)=3(2 - 1)(2 - 5)=-9\lt0\). So \(f(x)\) is decreasing on \((1,5)\).
• For \(x\gt5\), let \(x = 6\). Then \(f^\prime(6)=3(6 - 1)(6 - 5)=15\gt0\). So \(f(x)\) is increasing on \((5,\infty)\).

Step4: Find relative maxima and minima

By the first - derivative test:

• Since \(f(x)\) changes from increasing to decreasing at \(x = 1\), \(f(x)\) has a relative maximum at \(x = 1\).
• Since \(f(x)\) changes from decreasing to increasing at \(x = 5\), \(f(x)\) has a relative minimum at \(x = 5\).

Answer:

1. \(f\) is increasing on the intervals \((-\infty,1)\cup(5,\infty)\)
2. \(f\) is decreasing on the intervals \((1,5)\)
3. The relative maxima of \(f\) occur at \(x = 1\)
4. The relative minima of \(f\) occur at \(x = 5\)