Lesson Explainer: Critical Points and Local Extrema of a Function Mathematics • Higher Education

In this explainer, we will learn how to find critical points of a function and check for local extrema using the first derivative test.

A local maximum or local minimum can be defined independently of derivatives.

Definition: Local Maximum and Minimum

Suppose a function 𝑓 is defined on some interval and consider the local extrema at π‘₯=π‘Ž. For sufficiently small 𝛿>0 and for all π‘₯ contained in the interval ]π‘Žβˆ’π›Ώ,π‘Ž+𝛿[,

  • if 𝑓(π‘₯)≀𝑓(π‘Ž), then π‘₯=π‘Ž is a local maximum;
  • if 𝑓(π‘₯)β‰₯𝑓(π‘Ž), then π‘₯=π‘Ž is a local minimum.

This is intuitive as it means that the value of the function at π‘₯=π‘Ž is the largest or smallest possible value for any π‘₯ in a sufficiently small neighborhood around π‘₯=π‘Ž, depending on whether you have a local maximum or a local minimum respectively. In practice, however, it is easier to determine local maximum and local minimum values by finding critical points of a function and classifying them using the first derivative test.

So, what is a critical point?

Definition: Critical Point

The point (π‘₯,𝑦) is called a critical point of 𝑦=𝑓(π‘₯) if π‘₯ is in the domain of the function 𝑓(π‘₯) and either 𝑓′(π‘₯)=0𝑓′(π‘₯).orisundefined

These points can be classified as local minima, local maxima, or points of inflection.

In other words, to determine the critical points of a function, we take the first derivative of the function, set it equal to zero, and solve for π‘₯. We should also check if there are any π‘₯ values in the domain of the function that make the first derivative undefined. We then substitute these values of π‘₯ into the function 𝑦=𝑓(π‘₯) in order to find the values of 𝑦 and hence the coordinates of the critical points (π‘₯,𝑦).

For a differentiable function when the first derivative is equal to zero, critical points are also known as stationary points. It is the point where the function is neither increasing or decreasing, hence the name stationary.

Consider the function 𝑓(π‘₯)=π‘₯βˆ’2π‘₯ defined on the set of real numbers ℝ. The derivative of this function is given by 𝑓′(π‘₯)=2π‘₯βˆ’2.

Since the first derivative is defined for all values of π‘₯, we only need to establish possible values of π‘₯ such that the first derivative is equal to zero. Therefore, we need to solve 2π‘₯βˆ’2=0π‘₯=1.

Thus, there is a critical point at π‘₯=1. We can evaluate the function at this point to find 𝑓(1)=(1)βˆ’2(1)=βˆ’1.

Therefore, (1,βˆ’1) is a critical point of this function.

Now let’s look at a situation where the critical point occurs when the first derivative is undefined rather than equal to zero. Suppose we have the function 𝑓(π‘₯)=√π‘₯=π‘₯.

The first derivative is given by 𝑓′(π‘₯)=12π‘₯=12√π‘₯,οŽͺ and, although this cannot be equal to zero, it is clearly undefined at π‘₯=0 since we would need to divide by zero. The function is, however, defined at this point since 𝑓(0)=0, so (0,0) is a critical point of this function.

In order to classify these as either local maxima, local minima, or points of inflection, we can use the first derivative test, which involves checking the sign of the first derivative for values immediately around the critical point.

The first derivative test states the following.

Definition: First Derivative Test

Suppose 𝑓 is a function defined on some interval containing the critical point π‘₯=π‘Ž. Then, for sufficiently small πœ–>0,

  1. if 𝑓′(π‘Žβˆ’πœ–)>0 and 𝑓′(π‘Ž+πœ–)<0 (i.e., 𝑓′(π‘₯) switches signs from positive to negative as it crosses π‘₯=π‘Ž), then π‘₯=π‘Ž is a local maximum;
  2. if 𝑓′(π‘Žβˆ’πœ–)<0 and 𝑓′(π‘Ž+πœ–)>0 (i.e., 𝑓′(π‘₯) switches signs from negative to positive as it crosses π‘₯=π‘Ž), then π‘₯=π‘Ž is a local minimum;
  3. if 𝑓′(π‘Žβˆ’πœ–)>0 and 𝑓′(π‘Ž+πœ–)>0 or 𝑓′(π‘Žβˆ’πœ–)<0 and 𝑓′(π‘Ž+πœ–)<0 (i.e., 𝑓′(π‘₯) does not change signs as it crosses π‘₯=π‘Ž), then π‘₯=π‘Ž is a point of inflection.

This is a consequence of the mean value theorem.

When dealing with local maxima and minima, the points at which the derivative changes signs are also called turning points; however, not all critical points are turning points.

If the function is defined on some interval [𝑝,π‘ž] and π‘₯=π‘Ž is the only critical point contained in this interval, then we would need to check the sign of the first derivative in the intervals for the π‘₯-coordinate, [𝑝,π‘Ž[, ]π‘Ž,π‘ž], to see if there is a sign change. If there are 𝑁 critical points π‘₯=π‘Ž,π‘Ž,…,π‘ŽοŠ§οŠ¨οŒ­ inside this interval with π‘Ž<π‘Ž<β‹―<π‘ŽοŠ§οŠ¨οŒ­, then we would need to check the sign of the first derivative in the intervals: [𝑝,π‘Ž[,]π‘Ž,π‘Ž[,]π‘Ž,π‘Ž[,…,]π‘Ž,π‘Ž[,]π‘Ž,π‘ž].

If the function is defined on the whole set of real numbers ℝ and π‘₯=π‘Ž is the only critical point, then we would need to check the sign of the derivative of some value in the intervals ]βˆ’βˆž,π‘Ž[, ]π‘Ž,∞[. Similarly, if there are 𝑁 critical points for this function, then we would need to check the sign of the first derivative in the intervals: ]βˆ’βˆž,π‘Ž[,]π‘Ž,π‘Ž[,]π‘Ž,π‘Ž[,…,]π‘Ž,π‘Ž[,]π‘Ž,∞[.

In fact, it is sufficient to check a particular test value from these intervals to determine the sign of the first derivative. This would indicate whether we have a local maximum, a local minimum, or a point of inflection.

Consider the function 𝑓(π‘₯)=π‘₯+32π‘₯βˆ’6π‘₯+1. We can find the critical points by setting the first derivative equal to zero: 𝑓′(π‘₯)=3π‘₯+3π‘₯βˆ’6=0.

Thus, we have to solve this quadratic equation in order to determine the critical points: 3π‘₯+3π‘₯βˆ’6=03ο€Ήπ‘₯+π‘₯βˆ’2=03(π‘₯βˆ’1)(π‘₯+2)=0.

Therefore, the critical points occur when π‘₯=1 and π‘₯=βˆ’2. We can also evaluate the function at these π‘₯-coordinates to give 𝑓(1)=(1)+32(1)βˆ’6(1)+1=βˆ’52,𝑓(βˆ’2)=(βˆ’2)+32(βˆ’2)βˆ’6(βˆ’2)+1=11.

Thus, the critical points are located at (βˆ’2,11) and ο€Ό1,βˆ’52.

In order to apply the first derivative test, we check the sign of the first derivative at test values for π‘₯ contained in the intervals ]βˆ’βˆž,βˆ’2[, ]βˆ’2,1[, and ]1,∞[, which we can pick as π‘₯=βˆ’3,0, and 2 respectively. If we evaluate the first derivative at these points, we have 𝑓′(βˆ’3)=3(βˆ’3)+3(βˆ’3)βˆ’6=12>0,𝑓′(0)=3(0)+3(0)βˆ’6=βˆ’6<0,𝑓′(2)=3(2)+3(2)βˆ’6=12>0.

We can summarize this in a table to show the sign of the first derivative as it passes through the critical points:

π‘₯βˆ’3βˆ’20
𝑓′(π‘₯)> 00< 0
π‘₯012
𝑓′(π‘₯)< 00> 0

We can see that the sign of the first derivative switches from positive to negative as it crosses π‘₯=βˆ’2, while the first derivative switches from negative to positive as it crosses π‘₯=1.

Therefore, we have a local maximum at (βˆ’2,11) and a local minimum at ο€Ό1,βˆ’52. There are no critical points that are points of inflection of the function.

Suppose we had the same function but now defined in the interval [0,5]; we would take the same steps in determining the critical value but we would ignore any that fall outside this interval. For our function, we would only consider the critical value at π‘₯=1, which lies within the interval, and for the first derivative test, we would only need to check the sign of the first derivative at test values for π‘₯ located at the intervals [0,1[ and ]1,5]. This would give us the same result as the test values we chose for the function defined on the set of real numbers ℝ, π‘₯=0 and π‘₯=2, already fall within these intervals.

We will now look at a few examples to practice and deepen our understanding. For the first example, we only need to determine the critical values for a function defined on an interval and not classify them.

Example 1: Finding the Critical Point of a Cubic Function in a Given Interval

Determine the critical points of the function 𝑦=βˆ’8π‘₯ in the interval [βˆ’2,1].

Answer

In this example, we have to find the critical points (π‘₯,𝑦) of a cubic polynomial function defined on a particular interval.

Since polynomials are continuous and differentiable over their entire domain, we only need to find the values of π‘₯ such that the first derivative of 𝑦 is equal to zero in order to determine the critical points: dd𝑦π‘₯=βˆ’24π‘₯=0π‘₯=0.

Therefore, π‘₯=0 is a critical point as it lies within the interval [βˆ’2,1]. We can find the value of 𝑦 by substituting this π‘₯ value into the function to give 𝑦=βˆ’8(0)=0.

Hence, the critical point is located at (0,0).

In our next example, we will need to find the critical points and classify them as local minimum or maximum values, using the first derivative test, for a polynomial function defined on the whole set of real numbers ℝ.

Example 2: Finding the Local Maximum and Minimum Values of a Polynomial Function and the Values for π‘₯ Where They Occur

Determine, if any, the local maximum and minimum values of 𝑓(π‘₯)=βˆ’2π‘₯βˆ’9π‘₯βˆ’12π‘₯βˆ’15, together with where they occur.

Answer

In this example, we have to find the critical points (π‘₯,𝑓(π‘₯)) of a cubic polynomial function defined on the whole real line. In particular, we have to find those points that correspond to local maximum and minimum values. Since polynomials are continuous and differentiable over their entire domain, we do not need to worry about the derivative being undefined. We begin by first taking the derivative of the function 𝑓(π‘₯) and setting this equal to zero: 𝑓′(π‘₯)=βˆ’6π‘₯βˆ’18π‘₯βˆ’12=0.

Therefore, in order to determine the critical points, we need to solve this quadratic equation: βˆ’6π‘₯βˆ’18π‘₯βˆ’12=0βˆ’6ο€Ήπ‘₯+3π‘₯+2=0βˆ’6(π‘₯+2)(π‘₯+1)=0.

Therefore, the critical points occur when π‘₯=βˆ’1 and π‘₯=βˆ’2. We can substitute these π‘₯ values back into the function to find 𝑓(βˆ’1)=βˆ’2(βˆ’1)βˆ’9(βˆ’1)βˆ’12(βˆ’1)βˆ’15=βˆ’10,𝑓(βˆ’2)=βˆ’2(βˆ’2)βˆ’9(βˆ’2)βˆ’12(βˆ’2)βˆ’15=βˆ’11.

Hence the critical points are located at (βˆ’1,βˆ’10) and (βˆ’2,βˆ’11).

In order to classify these points as either a local maximum or minimum, we can use the first derivative test and check the sign of the first derivative in the intervals for the π‘₯ values ]βˆ’βˆž,βˆ’2[, ]βˆ’2,βˆ’1[, and ]βˆ’1,∞[. We can pick the test values contained in these intervals as π‘₯=βˆ’3,βˆ’1.5, and 0 respectively. The first derivative evaluated at these points are 𝑓′(βˆ’3)=βˆ’6(βˆ’3)βˆ’18(βˆ’3)βˆ’12=βˆ’12<0,𝑓′(βˆ’1.5)=βˆ’6(βˆ’1.5)βˆ’18(βˆ’1.5)βˆ’12=1.5>0,𝑓′(0)=βˆ’6(0)βˆ’18(0)βˆ’12=βˆ’12<0.

We can summarize this in a table to show the sign of the first derivative as it passes through the critical points,

π‘₯βˆ’3βˆ’2βˆ’1.5
𝑓′(π‘₯)< 00> 0
π‘₯βˆ’1.5βˆ’10
𝑓′(π‘₯)> 00< 0

We can see that the sign of the first derivative switches from negative (<0) to positive (>0) as it crosses π‘₯=βˆ’2, so the critical point at (βˆ’2,βˆ’11) is a local minimum by the first derivative test. The first derivative also switches from positive to negative as it crosses π‘₯=βˆ’1, so the critical point at (βˆ’1,βˆ’10) is a local maximum.

We note that we could have also arrived at this conclusion directly from the function evaluated at these points or the 𝑦-coordinates, since the polynomial is continuous and they are the only two critical points.

Therefore, one must be a maximum and the other a minimum; we can just compare the function values evaluated at these points, 𝑓(βˆ’2)<𝑓(βˆ’1), which tells us that (βˆ’2,βˆ’11) is a minimum and (βˆ’1,βˆ’10) is a maximum.

Therefore, the local maximum is βˆ’10 at π‘₯=βˆ’1, and the local minimum is βˆ’11 at π‘₯=βˆ’2.

This example is similar to the previous one but instead of finding the critical values, we are given one and have to determine the unknown coefficients appearing in a polynomial.

Example 3: Finding the Unknown Coefficients in a Quadratic Function given the Minimum Value of the Function

Given that the function 𝑓(π‘₯)=π‘₯+𝐿π‘₯+π‘€οŠ¨ has a minimum value of 2 at π‘₯=βˆ’1, determine the values of 𝐿 and 𝑀.

Answer

In this example, we have to use the knowledge of the value of a critical point, in particular a local minimum, to determine the unknown coefficients in a quadratic equation.

Since 𝐿 and 𝑀 are constants, we can apply the usual rules for differentiating power terms and calculate the first derivative of the function as 𝑓′(π‘₯)=2π‘₯+𝐿.

Since π‘₯=βˆ’1 is a critical value, the first derivative should vanish at this point. This allows us to determine the coefficient 𝐿: 𝑓′(βˆ’1)=2(βˆ’1)+𝐿=0𝐿=2.

As we are told in the question that the minimum value of 2 occurs at π‘₯=βˆ’1, we can use 𝑓(βˆ’1)=2 to determine the coefficient 𝑀: 𝑓(βˆ’1)=(βˆ’1)+2(βˆ’1)+𝑀=2𝑀=3.

For completeness, we can also use the first derivative test to check if this point is indeed a local minimum. We pick a test value π‘₯=βˆ’2 from the interval ]βˆ’βˆž,βˆ’1[ and π‘₯=0 from the interval ]βˆ’1,∞[. At these points, 𝑓′(βˆ’2)=2(βˆ’2)+2=βˆ’2<0,𝑓′(0)=2(0)+2=2>0.

π‘₯βˆ’2βˆ’10
𝑓′(π‘₯)< 00> 0

We can see that the sign of the first derivative switches from negative to positive as it crosses π‘₯=βˆ’1; therefore, it is indeed a local minimum.

In fact, a quadratic curve will only ever have one stationary point, the nature of which is determined by the sign of the leading coefficient (the π‘₯ term) of the function. The stationary point is a minimum if the leading coefficient is positive and a maximum if the leading coefficient is negative, due to the shape of the curve.

In this case, 𝑓(π‘₯) has a positive leading coefficient and therefore we can be sure, even without the first derivative test, that it will have a local minimum.

To summarize, the values of 𝐿 and 𝑀 are given by 𝐿=2,𝑀=3.

In our next example, we will demonstrate how to find a local maximum value for a function using the product rule with exponentials.

Example 4: Finding the Local Maximum Value of a Function Involving Using the Product Rule with Exponential Functions

Determine where 𝑓(π‘₯)=3π‘₯π‘’οŠ¨οŠ±ο— has a local maximum, and give the value there.

Answer

In this example, we have to find the local maximum of a function involving the product of an exponential and quadratic function.

Let’s begin by taking the first derivative. Recall that the product rule for two differentiable functions 𝑒 and 𝑣 is given by (𝑒𝑣)β€²=𝑒𝑣′+𝑒′𝑣.

Letting 𝑒=3π‘₯ and 𝑣=π‘’οŠ±ο—, we find 𝑒′=6π‘₯ and 𝑣′=βˆ’π‘’οŠ±ο—; therefore, 𝑓′(π‘₯)=ο€Ή3π‘₯(βˆ’π‘’)+(6π‘₯)(𝑒)=3π‘₯𝑒(2βˆ’π‘₯).οŠ¨οŠ±ο—οŠ±ο—οŠ±ο—

This derivative is always defined for π‘₯βˆˆβ„; therefore, we only have to consider the critical points when 𝑓′(π‘₯)=0 and we have to solve 3π‘₯𝑒(2βˆ’π‘₯)=0.οŠ±ο—

Since π‘’οŠ±ο— cannot be zero, we can multiply this expression by 𝑒 to find, 3π‘₯(2βˆ’π‘₯)=0.

Therefore the solutions to 𝑓′(π‘₯)=0 are given by π‘₯=0 and π‘₯=2, which are where the critical points occur. We can evaluate the function at these π‘₯-coordinates to give 𝑓(0)=3(0)𝑒=0,𝑓(2)=3(2)𝑒=12𝑒.()

Therefore the critical points are located at (0,0) and ο€Ή2,12π‘’ο…οŠ±οŠ¨. We can classify these points using the first derivative test by checking the signs of the first derivative at test values contained in the intervals ]βˆ’βˆž,0[, ]0,2[, and ]2,∞[, which we can choose as π‘₯=βˆ’1,1, and 3 respectively.

Evaluating the first derivative at these points, 𝑓′(βˆ’1)=3(βˆ’1)𝑒(2+1)=βˆ’9𝑒<0,𝑓′(1)=3(1)𝑒(2βˆ’1)=3𝑒>0,𝑓′(3)=3(3)𝑒(2βˆ’3)=βˆ’9𝑒<0.

We can summarize this in a table to show the sign of the first derivative as it passes through the critical points:

π‘₯βˆ’101
𝑓′(π‘₯)< 00> 0
π‘₯123
𝑓′(π‘₯)> 00< 0

We can see that the sign of the first derivative switches from negative to positive as it crosses π‘₯=0, so the critical point at (0,0) is a local minimum by the first derivative test. The first derivative also switches from positive to negative as it crosses π‘₯=2, so the critical point at ο€Ή2,12π‘’ο…οŠ±οŠ¨ is a local maximum.

We note that we could have also arrived at this conclusion directly from the function evaluated at these points or the 𝑦-coordinates, since the function is continuous and they are the only two critical points.

Therefore, one must be a maximum and the other a minimum; we can just compare the function values evaluated at these points, 𝑓(0)<𝑓(2), which tells us that (0,0) is a minimum and ο€Ή2,12π‘’ο…οŠ±οŠ¨ is a maximum.

Therefore, the location and value of the local maximum is given by π‘₯=2,𝑓(2)=12𝑒.

In our last example, we will learn how to determine any critical points for a function involving a logarithmic function and classify them using the first derivative test.

Example 5: Finding the Local Maximum and Minimum Values of a Function Involving a Logarithmic Function, If Any

Find, if any, the local maxima and minima for 𝑓(π‘₯)=3π‘₯βˆ’2π‘₯βˆ’4π‘₯ln.

Answer

In this example, we have to find any local maxima or minima of a function involving the sum of a logarithmic and quadratic function. Since the logarithmic function is defined only for positive values, the domain of the function is π‘₯>0.

Let’s begin by taking the first derivative of the function using the fact that: (π‘₯)β€²=1π‘₯ln: 𝑓′(π‘₯)=6π‘₯βˆ’2βˆ’4π‘₯=6π‘₯βˆ’2π‘₯βˆ’4π‘₯=2(3π‘₯+2)(π‘₯βˆ’1)π‘₯.

This derivative is clearly undefined at π‘₯=0 and is equal to zero at π‘₯=βˆ’23 and π‘₯=1. However, since the domain of 𝑓(π‘₯) is values of π‘₯ in the interval ]0,∞[ (i.e π‘₯>0), we only need to consider the point π‘₯=1. The function evaluated at this point is 𝑓(1)=3(1)βˆ’2(1)βˆ’41=1.ln

Therefore, the critical point is located at (1,1), and in order to classify this, we use the first derivative test. We need to check the sign of the first derivative at test values contained in the intervals for π‘₯<0: ]0,1[ and ]1,∞[, which we choose as π‘₯=0.5 and 2 respectively. We can evaluate the first derivative at these points to get 𝑓′(0.5)=6(0.5)βˆ’2βˆ’40.5=βˆ’7<0,𝑓′(2)=6(2)βˆ’2βˆ’42=8>0.

π‘₯0.512
𝑓′(π‘₯)< 00> 0

We can see that the sign of the first derivative switches from negative to positive as it crosses π‘₯=1.

Therefore, the local minimum equals 1 at π‘₯=1.

After establishing the critical points of a function, the second derivative test can also be used to determine whether we have local maximum or local minimum. In particular, if π‘₯=𝑐 is a critical point (i.e., 𝑓′(𝑐)=0 or is undefined), then the following is true:

  • if 𝑓′′(𝑐)<0, then 𝑓(π‘₯) has a local maximum at π‘₯=𝑐;
  • if 𝑓′′(𝑐)>0, then 𝑓(π‘₯) has a local minimum at π‘₯=𝑐;
  • if 𝑓′′(𝑐)=0, then 𝑓(π‘₯) may have a point of inflection at π‘₯=𝑐 (although this needs to be determined).

Points of inflection are where the function changes concavity. Concave up corresponds to a positive second derivative and concave down corresponds to a negative second derivative. Since the function changes from concave up to concave down (or vice versa), the second derivative must equal zero at that point. Thus, in order to determine whether an inflection point occurs when 𝑓′′(𝑐)=0, we consider the concavity on either side of the point π‘₯=𝑐.

However, this is beyond the scope of this explainer and will be covered elsewhere in more detail.

Key Points

  • At the critical points of a function 𝑓(π‘₯), we have 𝑓′(π‘₯)=0 or is undefined. We also need to check that these values are contained in the domain of the function.
  • A critical point is classified as either a local maximum, a local minimum, or a point of inflection.
  • The first derivative test can be used to classify critical points.

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