Lesson Explainer: Increasing and Decreasing Intervals of a Function Using Derivatives Mathematics • Higher Education

In this explainer, we will learn how to determine the increasing and decreasing intervals of functions using the first derivative of a function.

Derivatives are a fundamental tool of calculus. They can tell us more about a function than just the value of the slope at a given point or where the critical points of that function occur. The derivative has applications within classical mechanics, for instance, allowing us to determine information about the velocity or acceleration of an object given a differentiable function for its displacement.

Before we establish how to apply the derivative to find intervals of increase and decrease, letโ€™s begin by recapping the definitions of increasing and decreasing functions.

Definition: Increasing and Decreasing Functions on an Interval

A function ๐‘“ is said to be increasing on an interval ๐ผ if ๐‘“(๐‘ฅ)>๐‘“(๐‘ฅ)๐‘ฅ<๐‘ฅ๐ผ.๏Šจ๏Šง๏Šง๏Šจforallin

The function is said to be decreasing on ๐ผ if ๐‘“(๐‘ฅ)<๐‘“(๐‘ฅ)๐‘ฅ<๐‘ฅ๐ผ.๏Šจ๏Šง๏Šง๏Šจforallin

These definitions must be satisfied for every pair of points ๐‘ฅ๏Šง and ๐‘ฅ๏Šจ in ๐ผ with ๐‘ฅ<๐‘ฅ๏Šง๏Šจ.

Note

Since we have used the symbols โ€œ<โ€ and โ€œ>โ€ instead of โ€œโ‰คโ€ and โ€œโ‰ฅ,โ€ it can be said that the function is strictly increasing or decreasing on ๐ผ. Although, during this explainer, we will not be using this convention.

Consider the graph of ๐‘ฆ=๐‘“(๐‘ฅ), which is shown below.

The function has both increasing and decreasing intervals. We see the graph is increasing (in other words, slopes upward) for values of ๐‘ฅ<โˆ’2.5 and ๐‘ฅ>1 and is decreasing (slopes downward) for values of ๐‘ฅ such that โˆ’2.5<๐‘ฅ<1. The function is therefore increasing on the intervals ]โˆ’โˆž,โˆ’2.5[ and ]1,โˆž[ and decreasing on the interval ]โˆ’2.5,1[.

By considering the definition of the derivative of a function, we can define increasing and decreasing functions in an alternative way. The slope of the tangent to a curve at a given point is given by the derivative of the function at that point. For this reason, we can use calculus to determine whether a function is increasing or decreasing on ๐ผ.

Theorem: Increasing and Decreasing Functions Using Derivatives

Consider a function ๐‘ฆ=๐‘“(๐‘ฅ) that is differentiable on ]๐‘Ž,๐‘[.

If ๐‘“โ€ฒ(๐‘ฅ)>0 for all ๐‘ฅโˆˆ]๐‘Ž,๐‘[, then ๐‘“ is increasing on the interval ]๐‘Ž,๐‘[.

If ๐‘“โ€ฒ(๐‘ฅ)<0 for all ๐‘ฅโˆˆ]๐‘Ž,๐‘[, then ๐‘“ is decreasing on the interval ]๐‘Ž,๐‘[.

To prove this theorem, we let ๐‘ฅ๏Šง and ๐‘ฅโˆˆ]๐‘Ž,๐‘[๏Šจ, with ๐‘ฅ<๐‘ฅ๏Šง๏Šจ. Then, ๐‘“(๐‘ฅ) is differentiable on ]๐‘ฅ,๐‘ฅ[๏Šง๏Šจ. By the mean value theorem, there exists a number ๐‘โˆˆ]๐‘ฅ,๐‘ฅ[๏Šง๏Šจ such that ๐‘“(๐‘ฅ)โˆ’๐‘“(๐‘ฅ)=๐‘“โ€ฒ(๐‘)(๐‘ฅโˆ’๐‘ฅ).๏Šจ๏Šง๏Šจ๏Šง

For the first statement, if ๐‘“โ€ฒ(๐‘ฅ)>0 for ๐‘ฅโˆˆ]๐‘ฅ,๐‘ฅ[๏Šง๏Šจ, then ๐‘“โ€ฒ(๐‘)>0 since ๐‘ is in ]๐‘ฅ,๐‘ฅ[๏Šง๏Šจ.

Similarly, since ๐‘ฅ>๐‘ฅ๏Šจ๏Šง, we can say that ๐‘ฅโˆ’๐‘ฅ>0๏Šจ๏Šง.

The product of two positive numbers is positive, so the right-hand side of the equation is also positive.

This means that ๐‘“(๐‘ฅ)โˆ’๐‘“(๐‘ฅ)>0๏Šจ๏Šง, so ๐‘“(๐‘ฅ)>๐‘“(๐‘ฅ)๏Šจ๏Šง.

By the definition of increasing and decreasing functions, we can say that if ๐‘“โ€ฒ(๐‘ฅ)>0 on ]๐‘ฅ,๐‘ฅ[๏Šง๏Šจ, then ๐‘“ is an increasing function on the interval ]๐‘ฅ,๐‘ฅ[๏Šง๏Šจ.

We can prove the second part of this theorem in a similar way.

Now that we have the theorem, we can consider two examples that involve finding the intervals of increase and decrease of polynomial functions.

Example 1: Finding Intervals Where a Polynomial Function Increases and Decreases

Determine the intervals on which ๐‘“(๐‘ฅ)=๐‘ฅ2โˆ’4๐‘ฅ+2๏Šช๏Šจ is increasing or decreasing.

Answer

We know that for a differentiable function ๐‘ฆ=๐‘“(๐‘ฅ), ๐‘“ is said to be increasing on intervals where ๐‘“โ€ฒ(๐‘ฅ)>0 and decreasing on intervals where ๐‘“โ€ฒ(๐‘ฅ)<0.

Since ๐‘“ is a polynomial, it is differentiable everywhere. This means we can determine the intervals over which it is increasing and decreasing from its derivative. Using the power rule for differentiation, ๐‘“โ€ฒ(๐‘ฅ)=4๐‘ฅ2โˆ’2ร—4๐‘ฅ+0=2๐‘ฅโˆ’8๐‘ฅ.๏Šช๏Šฑ๏Šง(๏Šจ๏Šฑ๏Šง)๏Šฉ

The function will therefore be decreasing for values of ๐‘ฅ such that ๐‘“โ€ฒ(๐‘ฅ)<0 and increasing for values of ๐‘ฅ such that ๐‘“โ€ฒ(๐‘ฅ)>0.

To find such values of ๐‘ฅ, we begin by finding the values of ๐‘ฅ where ๐‘“โ€ฒ(๐‘ฅ)=0 by solving the equation 2๐‘ฅ๏€น๐‘ฅโˆ’4๏…=0๏Šจ. This gives us ๐‘ฅ=0, ๐‘ฅ=โˆ’2, and ๐‘ฅ=2. We then determine the sign of the first derivative over the intervals ]โˆ’โˆž,โˆ’2[, ]โˆ’2,0[, ]0,2[, and ]2,โˆž[ by substituting a test value from each into the expression 2๐‘ฅ๏€น๐‘ฅโˆ’4๏…๏Šจ. We will choose ๐‘ฅ=โˆ’3, ๐‘ฅ=โˆ’1, ๐‘ฅ=1, and ๐‘ฅ=3.

๐‘ฅโˆ’3โˆ’113
2๐‘ฅ๏€น๐‘ฅโˆ’4๏…๏Šจโˆ’306โˆ’630
Increasing or Decreasing?DecreasingIncreasingDecreasingIncreasing

The function must be decreasing on any intervals where ๐‘“โ€ฒ(๐‘ฅ)<0. According to the table, these are the intervals ]โˆ’โˆž,โˆ’2[ and ]0,2[.

Similarly, ๐‘“(๐‘ฅ) is increasing for values of ๐‘ฅ such that ๐‘“โ€ฒ(๐‘ฅ)>0. These are the intervals ]โˆ’2,0[ and ]2,โˆž[.

The function is decreasing on the intervals ]โˆ’โˆž,โˆ’2[ and ]0,2[ and increasing on the intervals ]โˆ’2,0[ and ]2,โˆž[.

In our first example, we saw how using a table to calculate test values for ๐‘ฅ contained in each interval can help us to identify the sign of the derivative. Since the endpoints of each interval are the values of ๐‘ฅ such that ๐‘“โ€ฒ(๐‘ฅ)=0, this test tells us what is happening to the graph of the function between each stationary point. In our second example, we will use the derivative to find intervals of increase and decrease of a quadratic function.

Example 2: Finding the Intervals of Increase and Decrease of a Quadratic Function

Determine the intervals on which the function ๐‘“(๐‘ฅ)=(โˆ’3๐‘ฅโˆ’12)๏Šจ is increasing and on which it is decreasing.

Answer

We know that for a function ๐‘ฆ=๐‘“(๐‘ฅ) that is differentiable on ]๐‘Ž,๐‘[, the following is true:

  • If ๐‘“โ€ฒ(๐‘ฅ)>0 for all ๐‘ฅโˆˆ]๐‘Ž,๐‘[, then ๐‘“ is increasing on the interval ]๐‘Ž,๐‘[.
  • If ๐‘“โ€ฒ(๐‘ฅ)<0 for all ๐‘ฅโˆˆ]๐‘Ž,๐‘[, then ๐‘“ is decreasing on the interval ]๐‘Ž,๐‘[.

Since ๐‘“(๐‘ฅ) is a polynomial, it is differentiable everywhere, so we can determine intervals of increase and decrease by considering the sign of its first derivative. Moreover, since ๐‘“(๐‘ฅ) is a differentiable function raised to a power, we can apply the general power rule to find its derivative.

This tells us that if ๐‘ฆ=[๐‘”(๐‘ฅ)]๏Š for a differentiable function ๐‘” and real constant ๐‘›, then ๐‘ฆโ€ฒ=๐‘›[๐‘”(๐‘ฅ)]๐‘”โ€ฒ(๐‘ฅ)๏Š๏Šฑ๏Šง. Applying this to ๐‘“(๐‘ฅ) gives ๐‘“โ€ฒ(๐‘ฅ)=2(โˆ’3๐‘ฅโˆ’12)ร—(โˆ’3)=โˆ’6(โˆ’3๐‘ฅโˆ’12)=18๐‘ฅ+72.๏Šจ๏Šฑ๏Šง๏Šง

The function is decreasing on any intervals where ๐‘“โ€ฒ(๐‘ฅ)<0. This is given by the following inequality: 18๐‘ฅ+72<0๐‘ฅ<โˆ’4.

Similarly, the function is increasing for values of ๐‘ฅ such that ๐‘“โ€ฒ(๐‘ฅ)>0: 18๐‘ฅ+72>0๐‘ฅ>โˆ’4.

The function is decreasing on the interval ]โˆ’โˆž,โˆ’4[ and increasing on the interval ]โˆ’4,โˆž[.

Note that there are two mathematical conventions about what to do with endpoints (in other words, whether to include a slope of 0 in the definition for increasing and decreasing functions). It is very much a matter of personal preference as to whether we choose to include these. In this explainer, we will be excluding the endpoints from our intervals.

In the first two examples, we calculated the intervals of increase and decrease for polynomial functions. It is important to realize that this process also holds for nonpolynomial functions such as logarithmic functions, exponential functions, trigonometric functions, and functions involving moduli, as demonstrated in our next example.

Example 3: Finding the Intervals of Increase and Decrease for a Modulus Function

Determine the intervals on which the function ๐‘“(๐‘ฅ)=(๐‘ฅ+3)|๐‘ฅ+3| is increasing and decreasing.

Answer

To establish intervals of increase and decrease for a function, we will begin by calculating its derivative, ๐‘“โ€ฒ(๐‘ฅ).If ๐‘“โ€ฒ(๐‘ฅ)>0 on an interval, the function is increasing over that interval. If ๐‘“โ€ฒ(๐‘ฅ)<0 on an interval, the function is decreasing over that interval.

Our function is the product of a linear function and the modulus of a linear function, so we will need to be careful in how we find its derivative. Letโ€™s think about the function |๐‘ฅ+3| as a piecewise function such that |๐‘ฅ+3|=๏ญโˆ’(๐‘ฅ+3)๐‘ฅ<โˆ’3,๐‘ฅ+3๐‘ฅโ‰ฅโˆ’3.forfor

We can use this definition to rewrite ๐‘“(๐‘ฅ) as follows: ๐‘“(๐‘ฅ)=๏ฐโˆ’(๐‘ฅ+3)๐‘ฅ<โˆ’3,(๐‘ฅ+3)๐‘ฅโ‰ฅโˆ’3.๏Šจ๏Šจforfor

Then, using the general power rule to differentiate gives ๐‘“โ€ฒ(๐‘ฅ)=๏ฎโˆ’2(๐‘ฅ+3)๐‘ฅ<โˆ’3,2(๐‘ฅ+3)๐‘ฅโ‰ฅโˆ’3.forfor

Note

This works because ๐‘“ is continuous at โˆ’3 and the left and right derivatives both exist and are equal to 0 at ๐‘ฅ=โˆ’3.

Since ๐‘“โ€ฒ(๐‘ฅ) is positive for all real ๐‘ฅโ‰ โˆ’3, ๐‘“ is increasing on โ„โˆ’{โˆ’3}. Since ๐‘“โ€ฒ(โˆ’3)=0, we can also include this value in the interval of increase if we choose. We deduce that ๐‘“(๐‘ฅ) is increasing over โ„.

In addition to polynomials and functions involving moduli, we can apply this process to logarithmic functions. We will demonstrate that in our next example.

Example 4: Finding the Intervals Where Functions Involving Logarithmic Functions Increase and Decrease

Given that ๐‘“(๐‘ฅ)=5๐‘ฅโˆ’3๐‘ฅโˆ’๐‘ฅ๏Šจln, find the intervals on which ๐‘“ is increasing or decreasing.

Answer

Remember, if ๐‘“ is differentiable on an open interval, then ๐‘“ will be increasing on intervals where ๐‘“โ€ฒ(๐‘ฅ)>0 and decreasing on intervals where ๐‘“โ€ฒ(๐‘ฅ)<0. So, we begin by finding ๐‘“โ€ฒ(๐‘ฅ), observing the domain of ๐‘“(๐‘ฅ) to be positive real numbers: ๐‘“โ€ฒ(๐‘ฅ)=10๐‘ฅโˆ’3โˆ’1๐‘ฅ.

To determine the sign of this function, we begin by finding the value of its ๐‘ฅ-intercepts by solving ๐‘“โ€ฒ(๐‘ฅ)=0: 10๐‘ฅโˆ’3โˆ’1๐‘ฅ=010๐‘ฅโˆ’3๐‘ฅโˆ’1=0(5๐‘ฅ+1)(2๐‘ฅโˆ’1)=0๐‘ฅ=โˆ’15๐‘ฅ=12.๏Šจor

Since โˆ’15 is outside of the domain of ๐‘“(๐‘ฅ), we will now determine the sign of ๐‘“โ€ฒ(๐‘ฅ) over the intervals ๏ 0,12๏” and ๏ 12,โˆž๏”. To do so, we substitute test values from each interval into the expression for ๐‘“โ€ฒ(๐‘ฅ). Letโ€™s choose ๐‘ฅ=110 and ๐‘ฅ=1.

๐‘ฅ1101
10๐‘ฅโˆ’3โˆ’1๐‘ฅโˆ’126
Increasing or DecreasingDecreasingIncreasing

Since ๐‘“โ€ฒ(๐‘ฅ)<0 at 110, it is decreasing here. Similarly, it is increasing at ๐‘ฅ=1.

๐‘“ is increasing on the interval ๏ 12,โˆž๏” and decreasing on the interval ๏ 0,12๏”.

Letโ€™s now demonstrate this process with a function that is the product of a polynomial and an exponential.

Example 5: Finding the Intervals of Increase and Decrease of a Function Using the Product Rule with Exponential Functions

Let ๐‘“(๐‘ฅ)=3๐‘ฅ๐‘’๏Šช๏Šฑ๏Šช๏—. Determine the intervals where this function is increasing and where it is decreasing.

Answer

We know that for a differentiable function ๐‘ฆ=๐‘“(๐‘ฅ), ๐‘“ will be increasing on intervals where ๐‘“โ€ฒ(๐‘ฅ)>0 and decreasing on intervals where ๐‘“โ€ฒ(๐‘ฅ)<0.

Letโ€™s begin by checking that the function ๐‘“(๐‘ฅ) is differentiable. We know that the product of two differentiable functions is also differentiable. 3๐‘ฅ๏Šช is a polynomial function and is therefore differentiable across its domain. Similarly, the exponential function ๐‘’๏Œบ๏— for real constants ๐‘Ž is also differentiable across its domain. This means that ๐‘“(๐‘ฅ) is indeed differentiable, so we can determine the intervals of increase and decrease from its derivative.

The product rule tells us that for two differentiable functions ๐‘ข and ๐‘ฃ, dddddd๐‘ฅ(๐‘ข๐‘ฃ)=๐‘ข๐‘ฃ๐‘ฅ+๐‘ฃ๐‘ข๐‘ฅ.

Let ๐‘ข=3๐‘ฅ๏Šช and ๐‘ฃ=๐‘’๏Šฑ๏Šช๏— so that ddanddd๐‘ข๐‘ฅ=12๐‘ฅ๐‘ฃ๐‘ฅ=โˆ’4๐‘’.๏Šฉ๏Šฑ๏Šช๏—

Substituting these values into the formula for the product rule gives ๐‘“โ€ฒ(๐‘ฅ)=3๐‘ฅร—๏€นโˆ’4๐‘’๏…+๐‘’ร—12๐‘ฅ=โˆ’12๐‘ฅ๐‘’+12๐‘ฅ๐‘’=12๐‘ฅ๐‘’(โˆ’๐‘ฅ+1).๏Šช๏Šฑ๏Šช๏—๏Šฑ๏Šช๏—๏Šฉ๏Šช๏Šฑ๏Šช๏—๏Šฉ๏Šฑ๏Šช๏—๏Šฉ๏Šฑ๏Šช๏—

The function will therefore be decreasing for values of ๐‘ฅ on the intervals where 12๐‘ฅ๐‘’(โˆ’๐‘ฅ+1)<0๏Šฉ๏Šฑ๏Šช๏— and increasing on the intervals where 12๐‘ฅ๐‘’(โˆ’๐‘ฅ+1)>0๏Šฉ๏Šฑ๏Šช๏—. Since ๐‘’>0๏Šฑ๏Šช๏— for all values of ๐‘ฅ, the intervals of increase and decrease will depend entirely on the sign of 12๐‘ฅ(โˆ’๐‘ฅ+1)๏Šฉ.

Letโ€™s calculate the stationary points of ๐‘“(๐‘ฅ) by solving 12๐‘ฅ(โˆ’๐‘ฅ+1)=0๏Šฉ. This gives us ๐‘ฅ=0 and ๐‘ฅ=1. Hence, we substitute values from the intervals ]โˆ’โˆž,0[, ]0,1[, and ]1,โˆž[ into the expression 12๐‘ฅ(โˆ’๐‘ฅ+1)๏Šฉ and establish their sign. We choose ๐‘ฅ=โˆ’1, ๐‘ฅ=0.5, and ๐‘ฅ=2.

๐‘ฅโˆ’10.52
12๐‘ฅ(โˆ’๐‘ฅ+1)๏Šฉโˆ’240.75โˆ’96
Increasing or Decreasing?DecreasingIncreasingDecreasing

๐‘“ is increasing when ๐‘“โ€ฒ(๐‘ฅ)>0 and decreasing when ๐‘“โ€ฒ(๐‘ฅ)<0. Hence, ๐‘“ is increasing on the interval ]0,1[ and decreasing on the intervals ]โˆ’โˆž,0[ and ]1,โˆž[.

In the previous examples, we looked at functions that are differentiable across their entire domain. This will not always be the case, so we will need to carefully consider the domain of the functions when finding intervals of increase and decrease.

Example 6: Finding the Intervals on Which a Function Involving a Root Function Is Increasing and Decreasing

Find the intervals on which the function ๐‘“(๐‘ฅ)=5๐‘ฅโˆšโˆ’5๐‘ฅ+3 is increasing and decreasing.

Answer

To establish intervals of increase and decrease for a function, we can consider its derivative, ๐‘“โ€ฒ(๐‘ฅ). If ๐‘“ is differentiable on an open interval, then ๐‘“ will be increasing on intervals where ๐‘“โ€ฒ(๐‘ฅ)>0 and decreasing on intervals where ๐‘“โ€ฒ(๐‘ฅ)<0.

Letโ€™s begin by checking that the function ๐‘“(๐‘ฅ) is differentiable. We know that the product of two differentiable functions is also differentiable. 5๐‘ฅ is a polynomial and therefore must be differentiable for all real numbers, though โˆšโˆ’5๐‘ฅ+3 is a little more complicated. It can be differentiated by applying the general power rule; however, we will need to make sure any values inside the square root are nonnegative: โˆ’5๐‘ฅ+3โ‰ฅ05๐‘ฅโ‰ค3๐‘ฅโ‰ค35.

We should note that technically, ๐‘“ is not differentiable at ๐‘ฅ=35 since we cannot take a right limit for its derivative at this point.

A function can only be increasing and decreasing on its domain, so we will only consider values of ๐‘ฅโ‰ค35.

To find an expression for ๐‘“โ€ฒ(๐‘ฅ), we can apply the product rule that tells us that for two differentiable functions ๐‘ข and ๐‘ฃ, dddddd๐‘ฅ(๐‘ข๐‘ฃ)=๐‘ข๐‘ฃ๐‘ฅ+๐‘ฃ๐‘ข๐‘ฅ.

Let ๐‘ข=5๐‘ฅ so that dd๐‘ข๐‘ฅ=5.

Similarly, let ๐‘ฃ=โˆšโˆ’5๐‘ฅ+3=(โˆ’5๐‘ฅ+3)๏Ž ๏Žก. Then, we apply the general power rule to find its derivative.

This tells us that if ๐‘ฆ=[๐‘”(๐‘ฅ)]๏Š for a differentiable function ๐‘”(๐‘ฅ), then ๐‘ฆโ€ฒ=๐‘›[๐‘”(๐‘ฅ)]๐‘”โ€ฒ(๐‘ฅ)๏Š๏Šฑ๏Šง, where ๐‘› is a real constant.

So, dd๐‘ฃ๐‘ฅ=12(โˆ’5๐‘ฅ+3)ร—(โˆ’5)=โˆ’52(โˆ’5๐‘ฅ+3).๏Žช๏Ž ๏Žก๏Žช๏Ž ๏Žก

By the product rule, ๐‘“โ€ฒ(๐‘ฅ)=5๐‘ฅร—๏€ฝโˆ’52(โˆ’5๐‘ฅ+3)๏‰+(โˆ’5๐‘ฅ+3)ร—5=โˆ’252๐‘ฅ(โˆ’5๐‘ฅ+3)+5(โˆ’5๐‘ฅ+3).๏Žช๏Ž ๏Žก๏Ž ๏Žก๏Žช๏Ž ๏Žก๏Ž ๏Žก

We can make this expression a little easier to work with by creating a common denominator of 2(โˆ’5๐‘ฅ+3)๏Ž ๏Žก and adding the expressions: ๐‘“โ€ฒ(๐‘ฅ)=โˆ’25๐‘ฅ2(โˆ’5๐‘ฅ+3)+5(โˆ’5๐‘ฅ+3)ร—2(โˆ’5๐‘ฅ+3)2(โˆ’5๐‘ฅ+3)=โˆ’25๐‘ฅ+10(โˆ’5๐‘ฅ+3)2(โˆ’5๐‘ฅ+3)=โˆ’75๐‘ฅ+302(โˆ’5๐‘ฅ+3)๐‘ฅ<35.๏Ž ๏Žก๏Ž ๏Žก๏Ž ๏Žก๏Ž ๏Žก๏Ž ๏Žก๏Ž ๏Žกfor

The function will therefore be decreasing when โˆ’75๐‘ฅ+302(โˆ’5๐‘ฅ+3)<0๏Ž ๏Žก and increasing when โˆ’75๐‘ฅ+302(โˆ’5๐‘ฅ+3)>0๏Ž ๏Žก for values of ๐‘ฅ<35.

The denominator of the fraction is positive for ๐‘ฅ<35, so the sign of the derivative is entirely determined by the sign of โˆ’75๐‘ฅ+30. Hence, the slope will be negative for all values of ๐‘ฅ in the domain of ๐‘“โ€ฒ where the following inequality is satisfied: โˆ’75๐‘ฅ+30<075๐‘ฅ>30๐‘ฅ>25.

This tells us that the function is decreasing for 25<๐‘ฅ<35.

Similarly, the function will be increasing when โˆ’75๐‘ฅ+30>0 and ๐‘ฅ<35: โˆ’75๐‘ฅ+30>075๐‘ฅ<3๐‘ฅ<25.

The function is increasing on ๏ โˆ’โˆž,25๏” and decreasing on ๏ 25,35๏”.

As we saw in the previous example, care must be taken when differentiating functions involving a root. This is also true for rational functions, where we will need to ensure the denominator of the function is not equal to zero. There will be occasions when the denominator cannot be equal to zero for any real values of the variable, in which case we can perform the calculations free from worry as we will see in the next example.

Example 7: Finding the Intervals of Increase and Decrease of a Rational Function

Determine the intervals on which the function ๐‘“(๐‘ฅ)=7๐‘ฅ๐‘ฅ+9๏Šจ is increasing and where it is decreasing.

Answer

To establish intervals of increase and decrease for a function, we can consider its derivative, ๐‘“โ€ฒ(๐‘ฅ). If ๐‘“ is differentiable on an open interval, then ๐‘“ is increasing on intervals where ๐‘“โ€ฒ(๐‘ฅ)>0 and decreasing on intervals where ๐‘“โ€ฒ(๐‘ฅ)<0.

The function ๐‘“(๐‘ฅ) is the quotient of two differentiable functions, so it is differentiable over its entire domain. We observe that there are no real values of ๐‘ฅ that make the denominator, ๐‘ฅ+9๏Šจ, equal to zero, so the domain is โ„.

To find the derivative, we use the quotient rule that says that for differentiable functions ๐‘ข and ๐‘ฃ, dd(๐‘ข๐‘ฃ)๐‘ฅ=๐‘ฃโˆ’๐‘ข๐‘ฃ.dddd๏‘๏—๏“๏—๏Šจ

We define ๐‘ข=7๐‘ฅ, so dd๐‘ข๐‘ฅ=7. Similarly, ๐‘ฃ=๐‘ฅ+9๏Šจ, so dd๐‘ฃ๐‘ฅ=2๐‘ฅ. Hence, ๐‘“โ€ฒ(๐‘ฅ)=7๏€น๐‘ฅ+9๏…โˆ’7๐‘ฅ(2๐‘ฅ)(๐‘ฅ+9)=โˆ’7๐‘ฅ+63(๐‘ฅ+9).๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

We might now observe that the denominator of the function ๐‘“โ€ฒ(๐‘ฅ) is ๏€น๐‘ฅ+9๏…๏Šจ๏Šจ, which is positive for all real values of ๐‘ฅ. This means that the sign of ๐‘“โ€ฒ(๐‘ฅ) is dependent entirely on the sign of the numerator.

To determine the sign of โˆ’7๐‘ฅ+63๏Šจ, we solve โˆ’7๐‘ฅ+63=0๏Šจ and then find the sign of test values in intervals around these points: โˆ’7๐‘ฅ+63=07๐‘ฅ=63๐‘ฅ=9๐‘ฅ=ยฑ3.๏Šจ๏Šจ๏Šจ

We will now choose test values of ๐‘ฅ from the intervals ]โˆ’โˆž,โˆ’3[, ]โˆ’3,3[, and ]3,โˆž[ and determine the sign of โˆ’7๐‘ฅ+63๏Šจ with these values, remembering that the function is increasing over intervals of ๐‘ฅ where ๐‘“โ€ฒ(๐‘ฅ)>0 and decreasing where ๐‘“โ€ฒ(๐‘ฅ)<0. Letโ€™s choose ๐‘ฅ=โˆ’4, ๐‘ฅ=0, and ๐‘ฅ=4.

๐‘ฅโˆ’404
โˆ’7๐‘ฅ+63๏Šจโˆ’4963โˆ’49
Increasing or Decreasing?DecreasingIncreasingDecreasing

Hence, the function is decreasing on the intervals ]โˆ’โˆž,โˆ’3[ and ]3,โˆž[ and increasing over ]โˆ’3,3[.

In our final example, we will demonstrate how to apply the process to find intervals of increase and decrease of trigonometric functions.

Example 8: Finding the Intervals of Increase and Decrease of a Function Involving Trigonometric Functions

For 0<๐‘ฅ<2๐œ‹5, find the intervals on which ๐‘“(๐‘ฅ)=5๐‘ฅ+35๐‘ฅcoscos๏Šจ is increasing or decreasing.

Answer

We can determine the intervals over which a differentiable function is increasing or decreasing by inspecting the sign of its first derivative.

Since ๐‘“(๐‘ฅ) is the sum of two cosine functions and the cosine function is differentiable across โ„, we can infer that ๐‘“(๐‘ฅ) is differentiable on โ„.

We know that for real constants ๐‘Ž, ddcossin๐‘ฅ๐‘Ž๐‘ฅ=โˆ’๐‘Ž๐‘Ž๐‘ฅ. We can therefore use a combination of this formula and the general power rule to differentiate each part of the function: ๐‘“โ€ฒ(๐‘ฅ)=2((5๐‘ฅ))ร—(โˆ’5(5๐‘ฅ))+5(โˆ’3(5๐‘ฅ))=โˆ’10(5๐‘ฅ)(5๐‘ฅ)โˆ’15(5๐‘ฅ)=โˆ’5(5๐‘ฅ)(2(5๐‘ฅ)+3).cossinsincossinsinsincos๏Šจ๏Šฑ๏Šง

The function will therefore be decreasing over intervals where โˆ’5(5๐‘ฅ)(2(5๐‘ฅ)+3)<0sincos and increasing over intervals where โˆ’5(5๐‘ฅ)(2(5๐‘ฅ)+3)>0sincos.

This means we need to determine the sign of โˆ’5(5๐‘ฅ)(2(5๐‘ฅ)+3)sincos at various points. To do so, we will find the ๐‘ฅ-intercepts of the graph of ๐‘“(๐‘ฅ)=โˆ’5(5๐‘ฅ)(2(5๐‘ฅ)+3)sincos by solving โˆ’5(5๐‘ฅ)(2(5๐‘ฅ)+3)=0sincos.

These occur when โˆ’5(5๐‘ฅ)=0sin or 2(5๐‘ฅ)+3=0cos.

So, โˆ’5(5๐‘ฅ)=0(5๐‘ฅ)=05๐‘ฅ=0+2๐œ‹๐‘›๐œ‹+2๐œ‹๐‘›๐‘›โˆˆโ„ค๐‘ฅ=0+25๐œ‹๐‘›๐œ‹5+25๐œ‹๐‘›๐‘›โˆˆโ„ค.sinsinorfororfor

In the interval 0<๐‘ฅ<2๐œ‹5, the only solution is when ๐‘›=0: ๐‘ฅ=๐œ‹5.

The process for solving 2(5๐‘ฅ)+3=0cos is as follows: 2(5๐‘ฅ)+3=02(5๐‘ฅ)=โˆ’3(5๐‘ฅ)=โˆ’32.coscoscos

There are no real solutions to this equation, so the only ๐‘ฅ-intercept of the graph of ๐‘“(๐‘ฅ)=โˆ’5(5๐‘ฅ)(2(5๐‘ฅ)+3)sincos is ๐‘ฅ=๐œ‹5. We will therefore choose test values from the intervals ๏Ÿ0,๐œ‹5๏“ and ๏ ๐œ‹5,2๐œ‹5๏” and check the sign of the derivative at these points. Letโ€™s choose ๐‘ฅ=๐œ‹10 and ๐‘ฅ=3๐œ‹10.

๐‘ฅ๐œ‹103๐œ‹10
โˆ’5(5๐‘ฅ)(2(5๐‘ฅ)+3)sincosโˆ’1515
Increasing or Decreasing?DecreasingIncreasing

Since the derivative is negative on the interval ๏Ÿ0,๐œ‹5๏“ and positive on the interval ๏ ๐œ‹5,2๐œ‹5๏”, the function is decreasing on ๏Ÿ0,๐œ‹5๏“ and increasing on ๏ ๐œ‹5,2๐œ‹5๏”.

We will now recap the key concepts from this explainer.

Key Points

  • A function ๐‘“ is increasing on an interval ๐ผ if ๐‘“(๐‘ฅ)<๐‘“(๐‘ฅ)๐‘ฅ<๐‘ฅ๐ผ๏Šง๏Šจ๏Šง๏Šจforallin and decreasing on ๐ผ if ๐‘“(๐‘ฅ)>๐‘“(๐‘ฅ)๐‘ฅ<๐‘ฅ๐ผ.๏Šง๏Šจ๏Šง๏Šจforallin
  • We can establish the intervals of increase and decrease for differentiable functions by considering the sign of the derivative.
    For a function ๐‘ฆ=๐‘“(๐‘ฅ) that is differentiable on ]๐‘Ž,๐‘[, the following is true:
    • If ๐‘“โ€ฒ(๐‘ฅ)>0, where ๐‘ฅโˆˆ]๐‘Ž,๐‘[, then ๐‘“ is strictly increasing on the interval ]๐‘Ž,๐‘[.
    • If ๐‘“โ€ฒ(๐‘ฅ)<0, where ๐‘ฅโˆˆ]๐‘Ž,๐‘[, then ๐‘“ is strictly decreasing on the interval ]๐‘Ž,๐‘[.

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