Using Key Idea 3.3.5, the domain of \(f\) is \(\mathbb{R}\) or \((-\infty,\infty)\text{.}\) Next, we find the critical values of \(f\text{.}\) We have \(\fp(x) = 3x^2+2x-1 = (3x-1)(x+1)\text{,}\) so \(\fp(x) = 0\) when \(x=-1\) and when \(x=1/3\text{.}\) \(\fp\) is never undefined.

We thus break the domain (in this case the real line) into three subintervals based on the two critical values we just found: \((-\infty,-1)\text{,}\) \((-1,1/3)\) and \((1/3,\infty)\text{.}\) This is shown in Figure 3.3.7.

We now pick a value \(p\) in each subinterval and find the sign of \(\fp(p)\text{.}\) All we care about is the sign, so we do not actually have to fully compute \(\fp(p)\text{;}\) pick “nice” values that make this simple.

Subinterval 1: \((-\infty,-1)\)

We (arbitrarily) pick \(p=-2\text{.}\) We can compute \(\fp(-2)\) directly: \(\fp(-2) = 3(-2)^2+2(-2)-1=7>0\text{.}\) We conclude that \(f\) is increasing on \((-\infty,-1)\text{.}\)

Note we can arrive at the same conclusion without computation. For instance, we could choose \(p=-100\text{.}\) The first term in \(\fp(-100)\text{,}\) i.e., \(3(-100)^2\) is clearly positive and very large. The other terms are small in comparison, so we know \(\fp(-100)>0\text{.}\) All we need is the sign.

Subinterval 2: \((-1,1/3)\)

We pick \(p=0\) since that value seems easy to deal with. \(\fp(0) = -1\lt 0\text{.}\) We conclude \(f\) is decreasing on \((-1,1/3)\text{.}\)

Subinterval 3: \((1/3,\infty)\)

Pick an arbitrarily large value for \(p>1/3\) and note that \(\fp(p) =3p^2+2p-1 >0\text{.}\) We conclude that \(f\) is increasing on \((1/3,\infty)\text{.}\)

Figure 3.3.8 summarizes our work.

We can verify our calculations by considering Figure 3.3.9, where \(f\) is graphed. The graph also presents \(\fp\text{;}\) note how \(\fp>0\) when \(f\) is increasing and \(\fp\lt 0\) when \(f\) is decreasing.

We start by noting the domain of \(f\text{:}\) \((-\infty,1)\cup(1,\infty)\text{.}\)

Since \(f\) is not defined at \(x=1\) (it has a vertical asymptote), the increasing/decreasing nature of \(f\) could switch at this value. We know that \(\fp(1)\) will be undefined since \(f\) is discontinuous at \(1\text{.}\) We do not formally consider \(x=1\) to be a critical value of \(f\text{,}\) but we will use \(1\) to subdivide the real number line.

Using the Quotient Rule 2.4.8, we find \begin{equation*} \fp(x) = \frac{x^2-2x-3}{(x-1)^2} \text{.} \end{equation*}

We need to find the critical values of \(f\text{;}\) we want to know when \(\fp(x)=0\) and when \(\fp\) is not defined. That latter is straightforward: when the denominator of \(\fp(x)\) is \(0\text{,}\) \(\fp\) is undefined. That occurs when \(x=1\text{,}\) which we've already recognized as an important value, but not a critical number.

\(\fp(x)=0\) when the numerator of \(\fp(x)\) is \(0\text{.}\) That occurs when \(x^2-2x-3 = (x-3)(x+1) = 0\text{;}\) i.e., when \(x=-1,3\text{.}\)

We have found that \(f\) has two critical numbers, \(x=-1,3\text{,}\) and at \(x=1\) something important might also happen. These three numbers divide the real number line into four subintervals: \begin{equation*} (-\infty,-1), (-1, 1), (1,3), \text{ and } (3,\infty). \end{equation*}

Pick a number \(p\) from each subinterval and test the sign of \(\fp\) at \(p\) to determine whether \(f\) is increasing or decreasing on that interval. Again, we do well to avoid complicated computations; notice that the denominator of \(\fp\) is always positive so we can ignore it during our work.

Interval 1: \((-\infty,-1)\)

Choosing a very small number (i.e., a negative number with a large magnitude) \(p\) returns \(p^2-2p-3\) in the numerator of \(\fp\text{;}\) that will be positive. Hence \(f\) is increasing on \((-\infty,-1)\text{.}\)

Interval 2: \((-1,1)\)

Choosing \(0\) seems simple: \(\fp(0)=-3\lt 0\text{.}\) We conclude \(f\) is decreasing on \((-1,1)\text{.}\)

Interval 3: \((1,3)\)

Choosing \(2\) seems simple: \(\fp(2) = -3\lt 0\text{.}\) Again, \(f\) is decreasing.

Interval 4: \((3,\infty)\)

Choosing an very large number \(p\) from this subinterval will give a positive numerator and (of course) a positive denominator. So \(f\) is increasing on \((3,\infty)\text{.}\)

In summary, \(f\) is increasing on the intervals \((-\infty,-1)\) and \((3,\infty)\) and is decreasing on the intervals \((-1,1)\) and \((1,3)\text{.}\) Since at \(x=-1\text{,}\) the sign of \(\fp\) switched from positive to negative, Theorem 3.3.10 states that \(f(-1)\) is a relative maximum of \(f\text{.}\) At \(x=3\text{,}\) the sign of \(\fp\) switched from negative to positive, meaning \(f(3)\) is a relative minimum. At \(x=1\text{,}\) \(f\) is not defined, so there is no relative extremum at \(x=1\text{.}\) As previously stated, \(x=1\) is a vertical asymptote of \(f\text{.}\)

This is summarized in the number line shown in Figure 3.3.13. Also, Figure 3.3.14 shows a graph of \(f\text{,}\) confirming our calculations. This figure also shows \(\fp\text{,}\) again demonstrating that \(f\) is increasing when \(\fp>0\) and decreasing when \(\fp\lt 0\text{.}\)

The domain of \(f\) is \(\mathbb{R}\) (you can take the odd root of both positive and negative nubmers). Next, we take the first derivative. Since we know we want to solve \(\fp(x) = 0\text{,}\) we will do some algebra after taking derivatives. \begin{align*} f(x) \amp = x^{\frac{8}{3}}-4x^{\frac{2}{3}}\\ \fp(x) \amp = \frac{8}{3} x^{\frac{5}{3}} - \frac{8}{3}x^{-\frac{1}{3}}\\ \amp = \frac{8}{3}x^{-\frac{1}{3}}\left(x^{\frac{6}{3}}-1\right)\\ \amp =\frac{8}{3}x^{-\frac{1}{3}}\left(x^2-1\right)\\ \amp =\frac{8}{3}x^{-\frac{1}{3}}(x-1)(x+1)\text{.} \end{align*}

This derivation of \(\fp\) shows that \(\fp(x) = 0\) when \(x=\pm 1\) and \(\fp\)is not defined when \(x=0\text{.}\) Thus we have three critical values, breaking the number line into four subintervals as shown in Figure .

Interval 1: \((\infty,-1)\)

We choose \(p=-2\text{;}\) we can easily verify that \(\fp(-2)\lt 0\text{.}\) So \(f\) is decreasing on \((-\infty,-1)\text{.}\)

Interval 2: \((-1,0)\)

Choose \(p=-1/2\text{.}\) Once more we practice finding the sign of \(\fp(p)\) without computing an actual value. We have \(\fp(p) = (8/3)p^{-1/3}(p-1)(p+1)\text{;}\) find the sign of each of the three terms at the chosen value of \(p\text{.}\) \begin{equation*} \fp(p) = \frac{8}{3} \cdot \underbrace{p^{-\frac{1}{3}}}_{\lt 0}\cdot \underbrace{(p-1)}_{\lt 0}\underbrace{(p+1)}_{>0}. \end{equation*}

We have a “negative × negative × positive” giving a positive number; \(f\) is increasing on \((-1,0)\text{.}\)

Interval 3: \((0,1)\)

We do a similar sign analysis as before, using \(p\) in \((0,1)\text{.}\) \begin{equation*} \fp(p) = \frac{8}{3} \cdot \underbrace{p^{-\frac{1}{3}}}_{>0}\cdot \underbrace{(p-1)}_{\lt 0}\underbrace{(p+1)}_{>0}. \end{equation*}

We have two positive factors and one negative factor; \(\fp(p)\lt 0\) and so \(f\) is decreasing on \((0,1)\text{.}\)

Interval 4: \((1,\infty)\)

Similar work to that done for the other three intervals shows that \(\fp(x)>0\) on \((1,\infty)\text{,}\) so \(f\) is increasing on this interval.

We conclude by stating that \(f\) is increasing on \((-1,0)\) and \((1,\infty)\) and decreasing on \((-\infty,-1)\) and \((0,1)\text{.}\) The sign of \fp changes from negative to positive around \(x=-1\) and \(x=1\text{,}\) meaning by Theorem 3.3.10 that \(f(-1)\) and \(f(1)\) are relative minima of \(f\text{.}\) As the sign of \(\fp\) changes from positive to negative at \(x=0\text{,}\) we have a relative maximum at \(f(0)\text{.}\) Figure 3.3.16 shows a graph of \(f\text{,}\) confirming our result. We also graph \(\fp\text{,}\) highlighting once more that \(f\) is increasing when \(\fp>0\) and is decreasing when \(\fp\lt 0\text{.}\)