Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 5

Applications of the Derivative

Intervals of Increase and Decrease

In this section, we discuss using the derivative to find the intervals where a function is increasing and decreasing. We must start by defining these terms.

Figure 5.7. Function decreasing on the interval ${\displaystyle \left(-\infty \text{, }0\right)}$ and increasing on the interval ${\displaystyle \left(0\text{, }e\right)}$

In Figure 5.7, above, we see that ${\displaystyle f\left(c\right)>f\left(d\right)}$ for any ${\displaystyle c<d}$ in the interval ${\displaystyle \left(-\infty \text{, }0\right)}$ and ${\displaystyle f\left(a\right)<f\left(b\right)}$ for any ${\displaystyle a<b}$ in the interval ${\displaystyle \left(0\text{, }e\right)}$. Thus, according to Definition 5.4, the function is decreasing on the interval ${\displaystyle \left(-\infty \text{, }0\right)}$ and increasing on the interval ${\displaystyle \left(0\text{, }e\right)}$, as we see on the graph. That is, the definition corresponds to our idea of increasing and decreasing.

When a function is increasing, as in Figure 5.8, below, the tangent lines are positive.

Figure 5.8. Increasing function showing tangent lines

When it is decreasing, as in Figure 5.9, below, then the slopes are negative.

Figure 5.9. Decreasing function showing tangent lines

The following theorem then seems plausible.

Theorem 5.5.

1. If ${\displaystyle f^{\prime }\left(x\right)>0}$ for any $x$ on the interval $J$, then the function $f$ is increasing on the interval $J$.
2. If ${\displaystyle f^{\prime }\left(x\right)<0}$ for any $x$ on the interval $J$, then the function $f$ is decreasing on the interval $J$.

Later, we will see how we can prove this theorem, for now we will learn to use it to find the intervals where a function is increasing and decreasing.

Let us think about this carefully. We need to be aware that when a continuous function changes from increasing to decreasing (or from decreasing to increasing) at some point $c$, it is because, at this point, either the tangent line is horizontal or there is no tangent line. See Figure 5.10, below.

Figure 5.10. Changing continuous function showing no tangent line (left and centre) and horizontal tangent line (right) at points of change

Thus, to find the intervals of increase and decrease, we must identify such numbers $c$, called “critical numbers.”

To identify the critical points, we must simplify and factor the derivative function as much as possible. Once we identify the critical points, we use Theorem 5.5 to decide if the function is increasing or decreasing near the critical point.

Example 5.24. Consider the function ${\displaystyle f\left(x\right)=\sqrt{x^2-9}}$ (see Example 5.1). To find the critical points, we use the derivative of this function,

${\displaystyle f^{\prime }\left(x\right)={\displaystyle \frac{x}{\sqrt{x^2-9}}}\text{.}}$

Then

${\displaystyle f^{\prime }\left(x\right)={\displaystyle \frac{x}{\sqrt{x^2-9}}}\ne 0}$

for all $x$ in its domain.

Since ${\displaystyle f^{\prime }\left(-3\right)}$ and ${\displaystyle f^{\prime }\left(3\right)}$ are undefined, the critical numbers are ${\displaystyle -3}$ and ${\displaystyle 3}$. We have two intervals to consider:

${\displaystyle \left(-\infty \text{, }-3\right)\text{and}\left(3\text{, }\infty \right)\text{.}\text{[Why?]}}$

Choose any number ${\displaystyle s}$ in the interval ${\displaystyle \left(-\infty \text{, }-3\right)}$, say ${\displaystyle -5}$, and see that for this number ${\displaystyle f^{\prime }\left(-5\right)<0}$; hence, ${\displaystyle f^{\prime }\left(x\right)<0}$ on the interval ${\displaystyle \left(-\infty \text{, }-3\right)}$.

In the same manner, we can see that ${\displaystyle f^{\prime }\left(x\right)>0}$ on the interval ${\displaystyle \left(3\text{, }\infty \right)}$.

By Theorem 5.5, the function is decreasing on ${\displaystyle \left(-\infty \text{, }-3\right)}$ and increasing on ${\displaystyle \left(3\text{, }\infty \right)}$.

Example 5.25. The domain of the function

${\displaystyle h\left(x\right)={\displaystyle \frac{4x-5}{4x^2-6x}}}$

of Example 5.2 is ${\displaystyle ℝ-\left\{0\text{, }3∕2\right\}}$.

Its derivative is

${\displaystyle h^{\prime }\left(x\right)={\displaystyle \frac{-2\left(8x^2-20x+15\right)}{{\left(4x^2-6x\right)}^2}}\text{,}}$

and since

${\displaystyle 8x^2-20x+15\ne 0\text{for all}x}$

(you may want to use the quadratic formula to see this), we conclude that $h$ does not have critical points in its domain.

Moreover,

${\displaystyle 8x^2-20x+15={\left(4x^2-6x\right)}^2>0\text{for all}x;}$

hence, ${\displaystyle h^{\prime }\left(x\right)<0}$ for all $x$ in its domain. By Theorem 5.5, the function is decreasing on its domain.

Sometimes it is necessary to make up a table to find the intervals of increase and decrease, as follows:

• in the first column, we indicate the intervals given by the critical numbers.
• in the next columns, we put all of the factors of the derivative that determine the sign (positive or negative) of the derivative.
• in the next column, we place the sign of the derivative according to the product of the signs of its factors.
• in the last column, we indicate the conclusion, whether the function is increasing or decreasing.

Example 5.26. To find the critical numbers of the function

${\displaystyle y=x^3-6x^2-15x+4\text{, }}$

we use Definition 5.6 above:

${\displaystyle y^{\prime }=3x^2-12x-15=3\left(x^2-4x-5\right)=3\left(x-5\right)\left(x+1\right)\text{.}}$

Solving, we find that ${\displaystyle y^{\prime }=3\left(x-5\right)\left(x+1\right)=0}$, and ${\displaystyle x=5}$ and ${\displaystyle x=-1}$. These critical numbers give three intervals: ${\displaystyle \left(-\infty \text{, }-1\right)}$, ${\displaystyle \left(-1\text{, }5\right)}$, and ${\displaystyle \left(5\text{, }\infty \right)}$.

Next, we make up a table as follows:

From this table, we conclude that the function is increasing on the intervals ${\displaystyle \left(-\infty \text{, }-1\right)}$ and ${\displaystyle \left(5\text{, }\infty \right)}$, and decreasing on ${\displaystyle \left(-1\text{, }5\right)}$.

Note: One way to find the sign in the table is to take any number in the indicated interval and test for the sign of each factor. For instance for $x=0$ in $-<!--\; -\; -->1<x<5$, we have $(x-<!--\; -\; -->5)=-<!--\; -\; -->5<0$ and $(x+1)=1>0$.

Example 5.27. The domain of the function ${\displaystyle y={\displaystyle \frac{x}{x^2+4x+3}}}$ is

${\displaystyle \left(-\infty \text{, }-3\right)\cup \left(-3\text{, }-1\right)\cup \left(-1\text{, }\infty \right)\text{.}}$

Its derivative is

${\displaystyle y^{\prime }={\displaystyle \frac{(x^2+4x+3)-x(2x+4)}{{(x+3)}^2{(x+1)}^2}}={\displaystyle \frac{-x^2+3}{{(x+3)}^2{(x+1)}^2}}={\displaystyle \frac{\left(\sqrt{3}-x\right)\left(\sqrt{3}+x\right)}{{(x+3)}^2{(x+1)}^2}}}$

The critical numbers are ${\displaystyle x=\sqrt{3}}$ and ${\displaystyle x=-\sqrt{3}}$.

From the domain and the critical numbers, we have five intervals to consider, as indicated in the table below. The sign of the derivative depends on the factors of the numerator, since the denominator is always positive; therefore, we must consider these factors on the table.

The function is increasing on the intervals ${\displaystyle (-\sqrt{3},-1)}$ and ${\displaystyle \left(-1\text{, }\sqrt{3}\right)}$; it is decreasing on the intervals $(-\infty ,-3)$, $(-3,-\sqrt{3})$ and $(\sqrt{3},\infty )$.