(O3) Intervals of Increase / Decrease#

By the end of the lesson you will be able to:

  • find the intervals of increase / decrease of a function.

Lecture Videos#

Slope of a Curve and the Derivative#

Increasing / Decreasing Test

For the interval \((a,b)\):

  • \(f'>0 \implies f\) is increasing

  • \(f'<0 \implies f\) is decreasing

How to find the intervals of increase / decrease.

  1. Calculate the derivative \(f'\)

  2. Find where \(f'(x)=0\) and \(f'\; \text{ DNE}\)

  3. Create a sign chart for \(f'\).

    • Use the \(x\)-values where \(f'(x)=0\) and \(f'\; \text{ DNE}\) to create the intervals for our sign chart.

  4. Determine the sign of \(f'\) on each interval.

    • Plug test numbers into the derivative.

Example 1#

Find the intervals on which \(f\) is increasing and those on which \(f\) is decreasing, for:

\[ f(x)=\tfrac{1}{3}x^3-2x^2+3x+1 \]

Click through the tabs to see the steps of our solution.

In this example, we are going to:

  1. Calculate the derivative \(f'\)

  2. Find where \(f'(x)=0\) and \(f'\; \text{ DNE}\)

  3. Create a sign chart for \(f'\).

  4. Determine the sign of \(f'\) on each interval.

The first step is to calculate the derivative (and simplify it):

\[ f'(x)= x^2-4x+3 = (x-3)(x-1) \]

We know that we will need to use this to find the critical numbers and sign chart, so we factor the derivative to make it easier to work with.

Next we need to determine where the derivative is equal to \(0\), which we do by setting up the equation and solving for \(x\).

\[\begin{split} f'(x) = 0 & \xrightarrow{\text{set up}} (x-3)(x-1)=0 \\ & \\ & \xrightarrow{\text{set each factor} =0} (x-3) =0 \quad \text{or}\quad (x-1)=0\\ & \\ & \xrightarrow{\text{solve for } x} x=3 \quad \text{or}\quad x=1\\ \end{split}\]

From here we see that \(f'(x)=0\) when \(x=3\) and \(x=1\). (This is easy to do, because we factored our derivative earlier.)

We also need to determine where the derivative is undefined. For a polynomial though, the derivative is always defined. So \(f'(x) \text{ DNE}\) does not happen for any \(x\)-values for this function.

Critical Numbers: Putting all of this together, we can say that this function has critical numbers \(x=3\) and \(x=1\). (Since these numbers are also in the domain of function \(f\).)

These \(x\)-values we just found, \(x=1\) and \(x=3\), split up the domain of \(f\) into separate intervals: \((-\infty,1)\), \((1,3)\), and \((3,\infty)\). (Note that the domain of \(f\) is all real numbers.)

We use these intervals to help create the sign chart for our derivative. In the first column we put the factors of our derivative and in the first row we list the intervals.

\[\begin{split} \begin{array}{cccc} & (-\infty,1) & (1,3) & (3,\infty)\\ \hline (x-1) & & & \\ \hline (x-3) & & & \\ \hline f' & & & \\ \hline f & & & \\ \hline \end{array} \end{split}\]

Next, we use a test number to determine the sign of each factor on each interval. To do this for the interval \((1,3)\) we would:

  1. pick a test number in that interval, say \(t=2\)

  2. plug the test number into each factor

  3. record if the result is positive or negative on the sign chart.

\[\begin{split} (x-1) &\xrightarrow{\text{plug-in } t=2} (2-1)=+\\ (x-3) &\xrightarrow{\text{plug-in } t=2} (2-3)=-\\ \end{split}\]

And then since \(f'(x)=(x-3)(x-1)\), we can conclude that on this interval \((1,3)\) the sign of \(f'\) would be \((-)\cdot(+)\) and therefore negative.

We repeat this process for all other intervals to get:

\[\begin{split} \begin{array}{cccc} & (-\infty,1) & (1,3) & (3,\infty)\\ \hline (x-1) & - & + & + \\ \hline (x-3) & - & - & + \\ \hline f' & + & - & + \\ \hline f & & & \\ \hline \end{array} \end{split}\]

Now that we know the intervals where the derivative \(f'\) is positive and negative, we use this to find the intervals where the original function \(f\) is increasing and decreasing. Remember:

  • \(f' + \implies f\) increasing

  • \(f' - \implies f\) decreasing

\[\begin{split} \begin{array}{cccc} & (-\infty,1) & (1,3) & (3,\infty)\\ \hline (x-1) & - & + & + \\ \hline (x-3) & - & - & + \\ \hline f' & + & - & + \\ \hline f & \text{inc} & \text{dec} & \text{inc} \\ \hline \end{array} \end{split}\]

Finally, we can conclude that:

  • \(f\) is increasing on the intervals \((-\infty,1)\) and \((3,\infty)\)

  • \(f\) is decreasing on the interval \((1,3)\)

Example 2#

Find the intervals on which \(f\) is increasing and those on which \(f\) is decreasing, for:

\[ f(x)=\dfrac{2x^2}{x^2-1} \]

Click through the tabs to see the steps of our solution.

In this example, we are going to:

  1. Calculate the derivative \(f'\)

  2. Find where \(f'(x)=0\) and \(f'\; \text{ DNE}\)

  3. Create a sign chart for \(f'\).

  4. Determine the sign of \(f'\) on each interval.

The first step is to calculate the derivative (and simplify it). Since the function is a fraction with variable terms on both the top and bottom, we use Quotient Rule:

\[\begin{split} \begin{aligned} f'(x)&= \underbrace{\dfrac{\big[ 2x^2 \big]' (x^2-1) - 2x^2 \big[ x^2-1 \big]' }{(x^2-1)^2}}_{\text{Quotient Rule}} \\ & \\ &= \dfrac{4x (x^2-1) - 2x^2 (2x) }{(x^2-1)^2} \\ & \\ &= \dfrac{4x^3 - 4x - 4x^3 }{(x^2-1)^2} \\ & \\ &= \dfrac{ - 4x }{(x^2-1)^2} \\ \end{aligned} \end{split}\]

In order to find where the derivative is equal to \(0\), we start with our derivative (written as a single fraction), set the numerator equal to \(0\) and solve for \(x\).

\[ f'(x) = 0 \xrightarrow{\text{numerator} =0} -4x=0 \]

From here we see that \(f'(x)=0\) when \(x=0\).

In order to find where the derivative is undefined, we set the denominator equal to \(0\) and solve for \(x\).

\[ f'(x) \text{ DNE} \xrightarrow{\text{denominator} =0} (x^2-1)^2=0 \]

From here we see that \(f'(x) \text{ DNE}\) when \(x=\pm 1\).

Critical Numbers: Putting all of this together, we can say that this function has a critical number at \(x=0\). Note that \(x=-1\) and \(x=1\) are not technically critical numbers because they are not in the domain of function \(f\).)

However, for the purposes of our sign chart we are going to consider all three of these \(x\)-values: \(x=-1\), \(x=0\), and \(x=1\).

These \(x\)-values we just found, \(x=-1\), \(x=0\), and \(x=1\), split up the domain of \(f\) into separate intervals: \((-\infty,-1)\), \((-1,0)\), \((0,1)\), and \((1,\infty)\). (Note that the domain of \(f\) is all real numbers except \(-1\) and \(+1\))

We use these intervals to help create the sign chart for our derivative. In the first column we put the factors of our derivative and in the first row we list the intervals.

\[\begin{split} \begin{array}{ccccc} & (-\infty,-1) & (-1,0) & (0,1) & (1,\infty) \\ \hline -4x & & & & \\ \hline (x^2-1)^2 & & & & \\ \hline f' & & & & \\ \hline f & & & & \\ \hline \end{array} \end{split}\]

Next, we use a test number to determine the sign of each factor on each interval. To do this for the interval \((-\infty,-1)\) we would:

  1. pick a test number in that interval, say \(t=-2\)

  2. plug the test number into each factor

  3. record if the result is positive or negative on the sign chart.

\[\begin{split} -4x &\xrightarrow{\text{plug-in } t=-2} -4(-2)=+\\ (x^2-1)^2 &\xrightarrow{\text{plug-in } t=-2} ((-2)^2-1)^2=+\\ \end{split}\]

And then since \(f'(x)=\dfrac{-4x}{(x^2-1)^2}\), we can conclude that on this interval \((-\infty,-1)\) the sign of \(f'\) would be \(\tfrac{(+)}{(+)}\) and therefore positive.

We repeat this process for all other intervals to get:

\[\begin{split} \begin{array}{ccccc} & (-\infty,-1) & (-1,0) & (0,1) & (1,\infty) \\ \hline -4x & + & + & - & - \\ \hline (x^2-1)^2 & + & + & + & + \\ \hline f' & + & + & - & - \\ \hline f & & & & \\ \hline \end{array} \end{split}\]

Now that we know the intervals where the derivative \(f'\) is positive and negative, we use this to find the intervals where the original function \(f\) is increasing and decreasing. Remember:

  • \(f' + \implies f\) increasing

  • \(f' - \implies f\) decreasing

\[\begin{split} \begin{array}{ccccc} & (-\infty,-1) & (-1,0) & (0,1) & (1,\infty) \\ \hline -4x & + & + & - & - \\ \hline (x^2-1)^2 & + & + & + & + \\ \hline f' & + & + & - & - \\ \hline f & \text{inc} & \text{inc} & \text{dec} & \text{dec} \\ \hline \end{array} \end{split}\]

Finally, we can conclude that:

  • \(f\) is increasing on the intervals \((-\infty,-1)\) and \((-1,0)\)

  • \(f\) is decreasing on the interval \((0,1)\) and \((1,\infty)\)