Question

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

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 $x=$

in the last two type the word
one\ if there is not a relative maximum or minimum.
note: you can earn partial credit on this problem.
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Explanation:

Step1: Find the first derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(f(x)=2 + x-x^{2}\), we have \(f^\prime(x)=1-2x\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(1-2x=0\). Solving for \(x\) gives \(x=\frac{1}{2}\).

Step3: Determine the intervals of increase and decrease

• Choose a test - point in the interval \((-\infty,\frac{1}{2})\), say \(x = 0\). Then \(f^\prime(0)=1-2\times0=1>0\). So \(f(x)\) is increasing on \((-\infty,\frac{1}{2})\).
• Choose a test - point in the interval \((\frac{1}{2},\infty)\), say \(x = 1\). Then \(f^\prime(1)=1-2\times1=-1<0\). So \(f(x)\) is decreasing on \((\frac{1}{2},\infty)\).

Step4: Find relative maxima and minima

Since \(f(x)\) changes from increasing to decreasing at \(x=\frac{1}{2}\), by the first - derivative test, \(f(x)\) has a relative maximum at \(x=\frac{1}{2}\) and no relative minimum (because the function only changes from increasing to decreasing, not the other way around).

Answer:

1. \((-\infty,\frac{1}{2})\)
2. \((\frac{1}{2},\infty)\)
3. \(\frac{1}{2}\)
4. none