We start by finding \(\fp(x)=3x^2-3\) and \(\fpp(x)=6x\text{.}\) To find the inflection points, we use Theorem 3.4.11 and find where \(\fpp(x)=0\) or where \(\fpp \)is undefined. We find \(\fpp \)is always defined, and is \(0\) only when \(x=0\text{.}\) So the point \((0,f(0))=(0,1)\) is the only possible point of inflection.

This possible inflection point divides the real line into two intervals, \((-\infty,0)\) and \((0,\infty)\text{.}\) We use a process similar to the one used in the previous section to determine increasing/decreasing. Pick any \(c\lt 0\text{;}\) \(\fpp(c)\lt 0\) so \(f\) is concave down on \((-\infty,0)\text{.}\) Pick any \(c>0\text{;}\) \(\fpp(c)>0\) so \(f\) is concave up on \((0,\infty)\text{.}\) Since the concavity changes at \(x=0\text{,}\) the point \((0,1)\) is an inflection point.

The number line in Figure 3.4.13 illustrates the process of determining concavity; Figure 3.4.14 shows a graph of \(f\) and \(\fpp\text{,}\) confirming our results. Notice how \(f\) is concave down precisely when \(\fpp(x)\lt 0\) and concave up when \(\fpp(x)>0\text{.}\)

We need to find \(\fp\)and \(\fpp\text{.}\) Using the Quotient Rule 2.4.8 and simplifying, we find \begin{align*} \fp(x)\amp=\frac{-(1+x^2)}{(x^2-1)^2}\amp\fpp(x)\amp= \frac{2x(x^2+3)}{(x^2-1)^3}\text{.} \end{align*}

To find the possible points of inflection, we seek to find where \(\fpp(x)=0\) and where \(\fpp\) is not defined. Solving \(\fpp(x)=0\) reduces to solving \(2x(x^2+3)=0\text{;}\) we find \(x=0\text{.}\) We find that \(\fpp \)is not defined when \(x=\pm 1\text{,}\) for then the denominator of \(\fpp \)is \(0\text{.}\) We also note that \(f\) itself is not defined at \(x=\pm1\text{,}\) having a domain of \((-\infty,-1)\cup(-1,1)\cup(1,\infty)\text{.}\) Since the domain of \(f\) is the union of three intervals, it makes sense that the concavity of \(f\) could switch across intervals. We technically cannot say that \(f\) has a point of inflection at \(x=\pm1\) as they are not part of the domain, but we must still consider these \(x\)-values to be important and will include them in our number line.

The important \(x\)-values at which concavity might switch are \(x=-1\text{,}\) \(x=0\) and \(x=1\text{,}\) which split the number line into four intervals as shown in Figure 3.4.16. We determine the concavity on each. Keep in mind that all we are concerned with is the sign of \(\fpp \)on the interval.

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

Select a number \(c\) in this interval with a large magnitude (for instance, \(c=-100\)). The denominator of \(\fp'(x)\) will be positive. In the numerator, the \(\left(c^2+3\right)\) factor will be positive and the \(2c\) factor will be negative. Thus the numerator is negative and \(\fpp(c)\) is negative. We conclude \(f\) is concave down on \((-\infty,-1)\text{.}\)

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

For any number \(c\) in this interval, the factor \(2c\) in the numerator will be negative, the factor \(\left(c^2+3\right)\) in the numerator will be positive, and the factor \(\left(c^2-1\right)^3\) in the denominator will be negative. Thus \(\fpp(c)>0\) and \(f\) is concave up on this interval.

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

Any number \(c\) in this interval will be positive and “small.” Thus the numerator is positive while the denominator is negative. Thus \(\fpp(c)\lt 0\) and \(f\) is concave down on this interval.

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

Choose a large value for \(c\text{.}\) It is evident that \(\fpp(c)>0\text{,}\) so we conclude that \(f\) is concave up on \((1,\infty)\text{.}\)

We conclude that \(f\) is concave up on \((-1,0)\) and on \((1,\infty)\) and concave down on \((-\infty,-1)\) and on \(\cup(0,1)\text{.}\) There is only one point of inflection, \((0,0)\text{,}\) as \(f\) is not defined at \(x=\pm 1\text{.}\) Our work is confirmed by the graph of \(f\) in Figure 3.4.17. Notice how \(f\) is concave up whenever \(\fpp\)is positive, and concave down when \(\fpp\)is negative. The inflection in \(f\) occurs where \(\fpp\) changes sign.

We want to maximize the rate of decrease, which is to say, we want to find where \(S'\) has a minimum. To do this, we find where \(S''\) is \(0\) and \(S''\) changes from negative to positive. We find \(S'(t)=4t^3-16t\) and \(S''(t)=12t^2-16\text{.}\) Setting \(S''(t)=0\) and solving, we get \(t=\sqrt{4/3}\approx 1.16\) (we ignore the negative solution for \(t\) since it does not lie in the domain of our function \(S\)).

Since \(S''(1)=-4\lt 0\) and \(S''(2)=32\gt 0\text{,}\) we can say \(S'(\sqrt{4/3})\) is a local minimum of \(S'\text{.}\) This is both the inflection point and the point of maximum decrease. This is the point at which things first start looking up for the company. After the inflection point, sales are still decreasing, but not decreasing quite as quickly as they had been.

A graph of \(S(t)\) and \(S'(t)\) is given in Figure 3.4.20. When \(S'(t)\lt 0\text{,}\) sales are decreasing; note how at \(t\approx 1.16\text{,}\) \(S'(t)\) is minimized. That is, sales are decreasing at the fastest rate at \(t\approx 1.16\text{.}\) On the interval of \((1.16,2)\text{,}\) \(S\) is decreasing but concave up, so the decline in sales is “leveling off.”

We find \(\fp(x)=-100/x^2+1\) and \(\fpp(x) = 200/x^3\text{.}\) We set \(\fp(x)=0\) and solve for \(x\) to find the critical values (note that \(\fp\) is not defined at \(x=0\text{,}\) but neither is \(f\) so this is not a critical value.) We find the critical values are \(x=\pm 10\text{.}\) We now evaluate the second derivative at these critical numbers. Evaluating \(\fpp(10)=0.1>0\text{,}\) so there is a local minimum at \(x=10\text{.}\) Evaluating \(\fpp(-10)=-0.1\lt 0\text{,}\) determining a relative maximum at \(x=-10\text{.}\) These results are confirmed in Figure 3.4.25.