# Applications of Differentiation – Inflection Points, Concavity, and the Second Derivative Test

In this section, we will show you how to use the second derivative test to find points of inflection and intervals of concavity of a function.

Before we explore the second derivative test, we need to understand what concavity and inflection points are.

Concavity

Concavity relates to the rate of change of a function’s derivative. A function is concave up (or upwards) where the derivative f’ is increasing, or when f’’ is positive.  A function is concave down where the derivative f’ is decreasing, or when f’’ is negative. Graphically, concave up looks like a ‘cup’; concave down looks like a ‘frown’. Concave up (looks like a cup). The slope of the tangent line is increasing. Concave down (looks like a frown). The slope of the tangent line is decreasing.

Inflection Points

An inflection point is the point where the concavity changes. Graphically, it is where the graph goes from concave up to concave down (and vice versa).

Let’s consider the example below: In order to find the inflection points graphically, let us first identify the concave up regions, or the ‘cups’, and concave down regions, or the ‘frowns’. The green represents the concave up region, or the ‘cups’. The yellow region represents the concave down region, or ‘frowns’. The point where the graph changes from concave up to concave down (and vice versa) is the inflection point, represented by the black dots in the figure above.

Now that we understand concavity and inflection points from a graphical standpoint, we can use the second derivative test to calculate the regions of concavity and inflection points.

The Second Derivative Test

Now that we have covered concavity and inflection points, we can connect both concepts together to apply the second derivative test:

 Definition: The Second Derivative Test Let f ” be the second derivative of function f on a given interval I = [a,b]. The graph of f is: concave up if f ”(x) > 0 on the given interval (f’’ is positive). concave down if f ”(x) < 0 on the given interval (f’’ is negative). Guidelines for Performing the Second Derivative Test Take the second derivative of the function. Find the value that will make the second derivative equal to zero or undefined. These values are called the inflection numbers. Set y’’ equal to zero and solve for the variable. if y’’ is a rational function, set it equal to zero to solve for the variable AND set the denominator equal to zero to see where y’’ is undefined. Plot the inflection numbers on a number line and substitute values from the number line into y’’ to identify regions of positive and negative concavity. Substitute the inflection numbers back inside the original equation F(x) to obtain the y coordinate of the inflection number.

Example 1:

If F(x) = 2x3 – 3x2 + 4x, find the following:

• F’’(x)
• Regions of concavity
• Inflection points

Solution:

Take the second derivative of the function.

First, take the first derivative using the power rule: F ‘’(x) = 12x – 6

Find the value that will make the second derivative equal to zero or undefined.

Now set F’’(x) = 0 to find the inflection numbers:

12x – 6 = 0

12x = 6

Divide each side by 12 to solve for x: Our only inflection number is x = ½.

Plot the inflection numbers on a number line

Our number line is below: Substitute values from the number line to identify regions of positive and negative concavity.

Let’s substitute a value inside F’’(x) that is between – and ½, such as zero.

F’’(0) = 12(0) – 6

F’’(0) = 0 – 6

F’’(0) = – 6

Since the result is negative, the concavity will be negative from – to ½.

Now let’s substitute a value inside F’’(x) that is between ½ and , such as 1.

F’’(1) = 12(1) – 6

F’’(1) = 12 – 6

F’’(1) = 6

Since the result is negative, the concavity will be positive from ½ to . Interval of positive concavity (concave up): (½, )

Interval of negative concavity (concave down): (-, ½)

Substitute the inflection numbers back inside the original equation F(x)

Now let’s find the inflection point by substituting the inflection number ( x = ½) into the original function F(x).

F(½) = 2(½)3 – 3(½)2 + 4(½) = 1.5 or 3/2

The inflection point is (½, 1.5)

Let’s graph F(x) to verify our results: As we can see, the region is indeed concave down from – to ½ and concave up from ½ to .

Example 2:

Find the regions of concavity of the function F(y) = 2y3 + 3y2 +1 as well as the inflection points:

Solution:

Take the second derivative of the function.

Firstly, let’s take the first derivative. F’’(y) = 12y + 6

Find the value that will make the second derivative equal to zero or undefined.

Now set F’’(y) equal to zero to find the inflection numbers:

0 = 12y + 6

subtract 6 from both sides:

-6 = 12y

Divide by 12 on both sides: Substitute values from the number line to identify regions of positive and negative concavity.

Find a number between – and -½ to substitute into F’’(x), such as -1:

F ‘’ (-1) = 12(-1) + 6

F’’(-1) =  -12 + 6

F’’(-1) =  -6

Because F’’(-1) is negative, the region between – and -½ is concave down.

Find a number between -½ and to substitute into F’’(x), such as 1:

F ‘’ (1) = 12(1) + 6

F’’(1) =  12 + 6

F’’(1) =  18

Because F’’(1) is positive, the region between -½ and is concave up.

Interval of positive concavity (concave up): (-½, )

Interval of negative concavity (concave down): (-, -½) Substitute the inflection numbers back inside the original equation F(x)

Now let’s find the inflection point by substituting the inflection number (x = -½) into the original function F(x).

F(-½) = 2(-½)3 + 3(-½)2 +1 = 1.5

The inflection point is (-½, 1.5)

Let’s graph F(x) to verify our results: The function is concave down from (-∞ , -1/2) and concave up from ( -½,  ∞).

Example 3: 0 = -3x2+4

3x2=4

Divide by 3 on both sides:

3x2/3 = 4/3

x2 = 4/3

Take the square root of both sides: So far, we have two inflection numbers, +1.155 and – 1.155.

Now we need to find the x values that make F’’(x) undefined. We see that F’’(x) is a rational function, so set the denominator equal to zero:

0 = (x2+4)3

Take the cube root of both sides to eliminate the cube on the right side:

(0)1/3 = [(x2+4)3]1/3

0 = x2+4

Subtract 4 from both sides:

-4 = x2

Take the square root of both sides: Because we have imaginary solutions, +2i and -2i, they will not be included in our list of inflection numbers.

Plot the inflection numbers on the number line: Substitute values from the number line into F’’(x) to identify intervals of positive and negative concavity

Let’s find a value between – and -1.155 to plug into F’’(x), such as -2: Since the result is positive, the concavity from 1.155 to is positive.

Intervals positive concavity (concave up): (-, -1.155) and (1.155, )

Intervals of negative concavity (concave down): (-1.155, 1.155) Substitute inflections numbers back into F(x) to find inflection points

To obtain the inflection points, plug in the inflection numbers, -1.155 and +1.155 into the original function F(x): The inflection points are (-1.155, 0.563) and (1.155, 0.563).

Let’s graph F(x) to verify our results: Indeed, the graph is concave up on the intervals (-∞, -1.155) and (1.155, ∞). The graph is concave down on the interval (-1.155, 1.155)

Example 4: Find the second derivative, inflection point, and intervals of concavity of F(x) = x4+2x+1

Solution:

Take the second derivative of the function

Using the power rule,

F’(x) = 4x3+2

F’’(x) = 12x2

Set F’’(x) equal to zero to find inflection numbers

12x2=0

x2= 0

x = 0

The only inflection number is x = 0

Plot the inflection numbers on the number line Substitute values from the number line into F’’(x) to identify intervals of positive and negative concavity

Let’s select a number between -∞ and 0, such as -1:

F’’(-1) = 12(-1)2= 12

Since F’’(-1) is positive, the graph is concave up from (-∞,0).

Let’s select a number between 0 and ∞, such as 1

F’’(1) = 12(1)2= 12

Since F’’(1) is positive, the graph is concave up from (0, ∞). Because the graph is concave up over the entire domain of F(x), there are no inflection points (if the concavity is the same over an interval [a,b], then there are no inflection points on that interval).

Let’s graph F(x) to confirm our findings: As you can see, the would-be inflection point of (0,1) is not an inflection point. The entire graph is concave up over the domain. Therefore, there is no change in concavity and therefore no inflection point exists.

 Summary of Section Concavity: Concavity relates to the rate of change of a function’s derivative. A function is concave up when f’’ is positive.  A function is concave down when f’’ is negative. Concave up = looks like a cup Concave down = looks like a frown Inflection Point: is a point on the graph where the concavity changes. Graphically, this can be identified when the graph changes from concave down to concave up (and vice versa). Analytically, this can be determined from the second derivative test. Second Derivative Test: Using the second derivative of a function to find inflection points and intervals of concavity. Find F’’(x). Set F’’(x) = 0 and solve to find inflection numbers. Create a sign chart to find intervals of positive and negative concavity. Substitute inflection numbers into F(x) to obtain the inflection point. Check your results by graphing.