4.5 Analyzing Slope
Let's get a little more practice analyzing slope using transition diagrams.
Example.
Consider the function
\[
\begin{align}
z &\depends x
\\
z &= x^3 - 3 x.
\end{align}
\]
By taking the derivative
\[
\begin{align}
\overset{x}{\diff} z &= \overset{x}{\diff} \bigl(x^3 - 3 x\bigr) = \overset{x}{\diff} \bigl(x^3\bigr) - \overset{x}{\diff} (3 x) = 3 x^2 \mult \overset{x}{\diff} x - 3 \mult \overset{x}{\diff} x = 3 x^2 - 3,
\end{align}
\]
we find our function's slope.
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= 3 x^2 - 3
\end{align}
\]
Let's draw a transition diagram for slope. We find level points by setting the slope equal to zero.
\[
\begin{align}
&\operatorname{\overset{\mathit{x}}{\slope}} z = 0
\\
&3 x^2 - 3 = 0
\\
&x^2 = 1
\\
&x = - 1 \orSpaced x = 1
\end{align}
\]
We'll mark both level points on an \( x \)-axis to begin drawing a transition diagram.
Between negative one and one, the slope is negative. At \( x = 0 \) for example, the slope is negative three.
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= 3 \mult 0^2 - 3 = -3
\end{align}
\]
As you can check, the slope is positive in each of the other regions.
We've found our transition diagram for slope!
Let's have one more example.
Example.
Now consider the function
\[
\begin{align}
z &\depends x
\\
z &= \frac{1}{x^2 - 1}.
\end{align}
\]
Let's take the derivative of \( z \).
\[
\begin{align}
\overset{x}{\diff} z &= \overset{x}{\diff} \biggl(\frac{1}{x^2 - 1}\biggr) = \frac{- 1}{\bigl(x^2 - 1\bigr)^2} \mult \overset{x}{\diff} \bigl(x^2 - 1\bigr) = \frac{- 1}{\bigl(x^2 - 1\bigr)^2} \mult \overset{x}{\diff} \bigl(x^2\bigr)
\\
&= \frac{- 1}{\bigl(x^2 - 1\bigr)^2} \mult 2 x \mult \overset{x}{\diff} x = \frac{- 2 x}{\bigl(x^2 - 1\bigr)^2}
\end{align}
\]
We've found the slope.
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= \frac{- 2 x}{\bigl(x^2 - 1\bigr)^2}
\end{align}
\]
The slope is zero at \( x = 0 \). The slope is undefined at \( x = -1 \) and at \( x = 1 \).
We can figure out where the slope is positive or negative by testing an \( x \)-value from each region.
The slope transitions at the level point, only.
| \( \vphantom{\overset{x}{o}} x \) | \( \operatorname{\overset{\mathit{x}}{\slope}} z \) |
|---|---|
| \( -2 \) | \( 4 / 9 \) |
| \( -1 \) | |
| \( -1 / 2 \) | \( 16 / 9 \) |
| \( 0 \) | \( 0 \) |
| \( 1 / 2 \) | \( - 16 / 9 \) |
| \( 1 \) | |
| \( 2 \) | \( - 4 / 9 \) |
For those that are curious, here are the graphs of the functions in our examples.
The first has a high point at \( x = -1 \) and a low point at \( x = 1 \). The second graph has a high point at \( x = 0 \) and no other level points.