(O4) Intervals of Concavity#

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

  • find the intervals of concavity of a function.

  • find all of its points of inflection.

Lecture Videos#

Finding Intervals of Concavity#

Second Derivative and Concavity

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

  • \(f''>0 \implies f\) is concave up

  • \(f''<0 \implies f\) is concave down

How to find the intervals of concavity.

  1. Calculate the second 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 of concavity and any inflection points, for:

\[ f(x)=x^4-4x^3 \]

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.

We begin by calculating the second derivative (and simplifying it):

\[\begin{split} \begin{aligned} f'(x) & = 4x^3-12x^2 \\ & \\ f''(x) & = 12x^2-24x =12x(x-2)\\ \end{aligned} \end{split}\]

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

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

\[\begin{split} f''(x) = 0 & \xrightarrow{\text{set up}} 12x(x-2)=0 \\ & \\ & \xrightarrow{\text{set each factor} =0} 12x =0 \quad \text{or}\quad (x-2)=0\\ & \\ & \xrightarrow{\text{solve for } x} x=0 \quad \text{or}\quad x=2\\ \end{split}\]

From here we see that \(f'(x)=0\) when \(x=0\) and \(x=2\). (This is easy to do, because we factored our derivative earlier.)

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

Possible Inflection Numbers: Putting all of this together, we can say that this function has possible inflection numbers \(x=0\) and \(x=2\).

These \(x\)-values we just found, \(x=0\) and \(x=2\), split up the domain of \(f\) into separate intervals: \((-\infty,0)\), \((0,2)\), and \((2,\infty)\). (Note that the domain of \(f\) is all real numbers.)

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

\[\begin{split} \begin{array}{cccc} & (-\infty,0) & (0,2) & (2,\infty)\\ \hline 12x & & & \\ \hline (x-2) & & & \\ \hline f'' & & & \\ \hline f & & & \\ \hline \end{array} \end{split}\]

Next, we use a test number to determine the sign of each factor on each interval. To do this for the interval \((-\infty,0)\) we would:

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

  2. plug the test number into each factor

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

\[\begin{split} \begin{aligned} 12x &\xrightarrow{\text{plug-in } t=-1} 12(-1)=-\\ (x-2) &\xrightarrow{\text{plug-in } t=-1} (-1-2)=-\\ \end{aligned} \end{split}\]

And then since \(f''(x)=12x(x-2)\), we can conclude that on this interval \((-\infty,0)\) the sign of \(f''\) would be \((-)\cdot(-)\) and therefore positive.

We repeat this process for all other intervals to get:

\[\begin{split} \begin{array}{cccc} & (-\infty,0) & (0,2) & (2,\infty)\\ \hline 12x & - & + & + \\ \hline (x-2) & - & - & + \\ \hline f'' & + & - & + \\ \hline f & & & \\ \hline \end{array} \end{split}\]

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

  • \(f'' + \implies f\) concave up

  • \(f'' - \implies f\) concave down

\[\begin{split} \begin{array}{cccc} & (-\infty,0) & (0,2) & (2,\infty)\\ \hline 12x & - & + & + \\ \hline (x-2) & - & - & + \\ \hline f'' & + & - & + \\ \hline f & \text{CU} & \text{CD} & \text{CU} \\ \hline \end{array} \end{split}\]

Finally, we can conclude that:

  • \(f\) is concave up on the intervals \((-\infty,0)\) and \((2,\infty)\)

  • \(f\) is concave down on the interval \((0,2)\)

Now that we know the intervals where \(f\) is concave up and concave down we are ready to identify the inflection numbers. Remember that we found possible inflection numbers: \(x=0\) and \(x=2\). In order for these to be actual inflection numbers:

  1. They need to be in the domain of \(f\).

  2. There needs to be a change in concavity.

Domain: Since \(f\) is a polynomial, both \(x\)-values are in the domain of \(f\). Check.

Change in Concavity: If we look at the last line of our sign chart we see that:

\[\begin{split} \begin{array}{cccc} & (-\infty,0) & (0,2) & (2,\infty)\\ \hline f & \text{CU} & \text{CD} & \text{CU} \\ \hline \end{array} \end{split}\]

  • at \(x=0\) the graph switches from CU to CD

  • at \(x=2\) the graph switches from CD to CU

So there is a change in concavity at both \(x\)-values.

Therefore, \(x=0\) and \(x=2\) are both inflection numbers since they satisfy the additional requirements.

Now that we know \(x=0\) and \(x=2\) are indeed inflection numbers, we need to find the inflection points. We do this by finding the corresponding \(y\)-coordinate for each point.

\[\begin{split} \begin{aligned} x=0 &\xrightarrow{\text{plug-in to } f} y_0=f(0)=(0)^4-4(0)^3 = 0\\ x=2 &\xrightarrow{\text{plug-in to } f} y_2=f(2)=(2)^4-4(2)^3 = -16\\ \end{aligned} \end{split}\]

Which means the graph of \(f\) has inflection points at:

\[ (0,0)\quad \text{and} \quad (2,-16) \]

Example 2#

Find the intervals of concavity and any inflection points, 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.

We begin by calculating the second derivative (and simplifying it). (Note we calculated the first derivative in a previous example

\[\begin{split} \begin{aligned} f'(x) & = \dfrac{-4x}{(x^2-1)^2} \\ & \\ f''(x) & = \dfrac{\big[ -4x \big]'(x^2-1)^2 - (-4x)\big[ (x^2-1)^2 \big]'}{\big((x^2-1)^2\big)^2}\\ & \\ & = \dfrac{-4(x^2-1)^2 +4x\cdot 2(x^2-1)\big[ x^2-1 \big]'}{(x^2-1)^4}\\ & \\ & = \dfrac{-4(x^2-1)^2 +8x(x^2-1)(2x)}{(x^2-1)^4}\\ & \\ & = \dfrac{(x^2-1)\cdot\big(-4(x^2-1) +8x(2x)\big)}{(x^2-1)^4}\\ & \\ & = \dfrac{-4(x^2-1) +8x(2x)}{(x^2-1)^3}\\ & \\ & = \dfrac{-4x^2+4 +16x^2}{(x^2-1)^3}\\ & \\ & = \dfrac{12x^2+4}{(x^2-1)^3}\\ \end{aligned} \end{split}\]

We know that we will need to use this to find the possible inflection numbers and sign chart, so we simplified the second derivative as much as possible to make it easier to work with.

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

\[ f''(x) = 0 \xrightarrow{\text{numerator} =0} 12x^2+4=0 \]

However, there is no solution for this equation. So NONE, there are no possible inflection numbers with \(f''(x)=0\).

To see that there are no solutions for this equation, you could try using the quadratic formula and you would end up with a negative under the square root, indicating that there are no real number solutions:

\[ 12x^2+4=0 \xrightarrow{\text{quadratic}} x=\dfrac{-0\pm\sqrt{(0)^2-4(12)(4)}}{2(12)} \]

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

\[ f''(x) \text{ DNE} \xrightarrow{\text{denominator} =0} (x^2-1)^3=0 \]

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

\[\begin{split} \begin{aligned} (x^2-1)^3 =0 & \implies x^2-1= 0 \\ & \implies x^2= 1 \implies x = \pm \sqrt{1} \\ \end{aligned} \end{split}\]

Possible Inflection Numbers: Putting all of this together, we can say that this function has possible inflection numbers \(x=-1\) and \(x=1\).

These \(x\)-values we just found, \(x=-1\) and \(x=1\), split up the domain of \(f\) into separate intervals: \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\).

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

\[\begin{split} \begin{array}{cccc} & (-\infty,-1) & (-1,1) & (1,\infty)\\ \hline 12x^2+4 & & & \\ \hline (x^2-1)^3 & & & \\ \hline f'' & & & \\ \hline f & & & \\ \hline \end{array} \end{split}\]

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

\[\begin{split} \begin{aligned} 12x^2+4 &\xrightarrow{\text{plug-in } t=-2} 12(-2)^2+4=+\\ (x^2-1)^3 &\xrightarrow{\text{plug-in } t=-2} ((-2)^2-1)^3=+\\ \end{aligned} \end{split}\]

And then since \(f''(x)=\tfrac{12x^2+4}{(x^2-1)^3}\), 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:

\[\begin{split} \begin{array}{cccc} & (-\infty,-1) & (-1,1) & (1,\infty)\\ \hline 12x^2+4 & + & + & + \\ \hline (x^2-1)^3 & + & - & + \\ \hline f'' & + & - & + \\ \hline f & & & \\ \hline \end{array} \end{split}\]

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

  • \(f'' + \implies f\) concave up

  • \(f'' - \implies f\) concave down

\[\begin{split} \begin{array}{cccc} & (-\infty,-1) & (-1,1) & (1,\infty)\\ \hline 12x^2+4 & + & + & + \\ \hline (x^2-1)^3 & + & - & + \\ \hline f'' & + & - & + \\ \hline f & \text{CU} & \text{CD} & \text{CU} \\ \hline \end{array} \end{split}\]

Finally, we can conclude that:

  • \(f\) is concave up on the intervals \((-\infty,-1)\) and \((1,\infty)\)

  • \(f\) is concave down on the interval \((-1,1)\)

Now that we know the intervals where \(f\) is concave up and concave down we are ready to identify the inflection numbers. Remember that we found possible inflection numbers: \(x=-1\) and \(x=1\). In order for these to be actual inflection numbers:

  1. They need to be in the domain of \(f\).

  2. There needs to be a change in concavity.

Domain: Since \(f\) is a rational function, we need to check where the denominator is equal to \(0\). And here we run into an issue:

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

So neither of these possible inflection numbers are in the domain of \(f\), which means we cannot have an inflection point at either of them.

Change in Concavity: If we look at the last line of our sign chart we see that:

\[\begin{split} \begin{array}{cccc} & (-\infty,-1) & (-1,1) & (1,\infty)\\ \hline f & \text{CU} & \text{CD} & \text{CU} \\ \hline \end{array} \end{split}\]

Even with these apparent changes in concavity, since \(x=-1\) and \(x=1\) are not in the domain of \(f\) they are not actual inflection numbers.

Our only two possible inflection numbers \(x=-1\) and \(x=1\) both turned out to not be actual inflection numbers. What does this mean?

It means this function does not have any inflection points.