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How To Find The Domain_3 And Period Of A Function

Written By Wednesday, 9 March 2022 Add Comment Edit

In this explainer, we will larn how to determine the domain and range of a trigonometric function.

Let us begin by recalling the definitions of the domain and range of a function.

Theorem: Domain and Range of a Function

The domain of a function 𝑓 ( π‘₯ ) is the set up of all possible values π‘₯ such that role 𝑓 ( π‘₯ ) is defined.

The range of a function 𝑓 ( π‘₯ ) is the set of all possible values the role 𝑓 ( π‘₯ ) can take, when π‘₯ is whatever number from the domain of the function.

In particular, we tin can find the domain and range of a function from its graph. Given a graph of a function, its domain is the part of the horizontal centrality where the graph exists, and its range is the part of the vertical axis where the graph exists.

One of the important characteristics of the graph of a trigonometric function is that the pattern of the graph repeats indefinitely. When the behavior of function 𝑓 ( π‘₯ ) repeats over an interval of length 𝐿 , then nosotros say that 𝑓 ( π‘₯ ) is periodic with period 𝐿 . In other words, a periodic function with menstruum 𝐿 must satisfy 𝑓 ( π‘₯ ) = 𝑓 ( π‘₯ + 𝐿 ) , for whatsoever π‘₯ in the domain.

To study the domain and the range of a periodic role, we commencement need to empathise the domain and the range of the function in an interval of length 𝐿 , for instance [ 0 , 𝐿 ] . We can choose whatsoever interval of length 𝐿 , but it will be user-friendly in most cases to consider the interval [ 0 , 𝐿 ] . Since the function is periodic with period 𝐿 , the function volition repeat the same behavior exterior of this interval. Hence, the range of the periodic function volition exist the same as its range over [ 0 , 𝐿 ] . Furthermore, if the part is defined for all values in [ 0 , 𝐿 ] , the function volition be defined for whatsoever values outside this interval every bit well due to its periodicity. In this instance, the domain of the periodic function is all real numbers, denoted ℝ or ] − ∞ , ∞ [ .

In our first example, we will decide the domain and the range of a periodic function from its given graph.

Example i: Determining the Domain and Range of a Function from the Graph

Consider the following graph of 𝑓 ( πœƒ ) .

  1. What is the domain of 𝑓 ( πœƒ ) ?
  2. What is the range of 𝑓 ( πœƒ ) ?

Answer

Before nosotros answer the questions almost the domain and range of 𝑓 ( πœƒ ) , we note that the graph of the part repeats indefinitely. This means that the part 𝑓 ( πœƒ ) is a periodic part. Nosotros tin run into that the function has a local minimum at πœƒ = 0 and returns to the same identify at πœƒ = 2 πœ‹ . From there, the office repeats the same values. This means that the menstruation of this function is ii πœ‹ .

Role 1

In this part, we demand to determine the domain of a periodic office from the given graph. We noted that the period of 𝑓 ( πœƒ ) is 2 πœ‹ . Think that, if a periodic part is divers within an interval whose length is equal to the catamenia, the function is defined for all real numbers. From the given graph, we can encounter that 𝑓 ( πœƒ ) is defined for all values within the interval [ 0 , 2 πœ‹ ] , whose length is equal to the catamenia ii πœ‹ .

Hence, the domain of 𝑓 ( πœƒ ) is all real numbers, or ] − ∞ , ∞ [ .

Office 2

In this part, we need to determine the range of a periodic function from the given graph. We noted that the menstruation of 𝑓 ( πœƒ ) is 2 πœ‹ . Call up that if the range of a periodic office is the aforementioned as the range of the function over an interval whose length is equal to the period, the function is defined for all real numbers. From the given graph, we can run into that the minimum value for the role 𝑓 ( πœƒ ) over the interval [ 0 , two πœ‹ ] is − four , and its maximum value over this interval is half dozen. Since 𝑓 ( πœƒ ) takes all values between its maximum and minimum, the range of 𝑓 ( πœƒ ) over this interval is [ − 4 , half-dozen ] .

Hence, the range of 𝑓 ( πœƒ ) is [ − 4 , vi ] .

In the previous example, we considered the domain and range of a periodic function from the given graph. Nosotros tin use the aforementioned method to find the domain and range of sine and cosine functions. Call back that the angle of 2 πœ‹ radians measures a total revolution on the unit circle. This means that the values of trigonometric ratios, sine and cosine, on the unit of measurement circumvolve would remain the same if we add two πœ‹ radians to any angle. This means that, for any angle of πœƒ radians, s i n s i n c o s c o south πœƒ = ( πœƒ + 2 πœ‹ ) , πœƒ = ( πœƒ + two πœ‹ ) .

This tells the states that the sine and cosine functions are periodic with a period of ii πœ‹ radians. Hence, we can find the domain and range of the sine and cosine functions by considering the graph of these functions over an interval of length ii πœ‹ . Consider the graphs of s i n πœƒ and c o s πœƒ .

Both graphs above are over the interval [ − three πœ‹ , three πœ‹ ] , just we only demand the graph over an interval of length 2 πœ‹ . So, we can consider this graph over the interval [ 0 , 2 πœ‹ ] to find the domain and range of these functions. Since both functions are defined everywhere in the interval [ 0 , 2 πœ‹ ] , we know that the domains of both the sine and cosine functions are all real numbers.

We tin can also note that the minimum value of both functions over the interval [ 0 , 2 πœ‹ ] is − ane , and the maximum value is i. Since both functions take all values between the maximum and minimum, the range of sine and cosine functions over this interval, and hence over their domains, is [ − 1 , 1 ] .

We summarize these results as follows.

Definition: Domain and Range of Sine and Cosine Functions

The domain of functions southward i n πœƒ and c o southward πœƒ is all real numbers, denoted either ] − ∞ , ∞ [ or ℝ .

The range of functions s i n πœƒ and c o s πœƒ is [ − 1 , i ] .

Above, we determined the domain and the range of s i north πœƒ and c o s πœƒ by using the graph and the periodicity of these trigonometric functions. In our adjacent example, we will use the aforementioned method to determine the domain and the range of a periodic function.

Instance 2: Finding Domain and Range of a Periodic Role from its Graph

The post-obit graph shows the role 𝑓 ( πœƒ ) . Assume the function has a period of two πœ‹ .

  1. What is the domain of 𝑓 ( πœƒ ) ?
  2. What is the range of 𝑓 ( πœƒ ) ?

Answer

Nosotros know that all characteristics of a periodic function are independent in the graph of a function over an interval whose length is equal to the period. This role has a menstruum of 2 πœ‹ ; hence we only need to consider its graph over the interval of length two πœ‹ . In this example, the function 𝑓 ( πœƒ ) is graphed over an interval with a length larger than 2 πœ‹ , and so this should provide sufficient information to determine the domain and the range of 𝑓 ( πœƒ ) .

Part 1

The domain of a function is a ready of all possible input values. From the given graph, we can see that the function is well defined at any values of πœƒ . Thus, the domain of 𝑓 ( πœƒ ) is all existent numbers, or ] − ∞ , ∞ [ .

Part 2

The range of a part is a set of all possible function values. From the given graph, nosotros detect that this function oscillates between − vii and 3, taking all values in between. The graph never goes beneath − seven or in a higher place 3 in the vertical axis. Hence, the range of 𝑓 ( πœƒ ) is [ − 7 , 3 ] .

When we are given algebraic expressions of trigonometric functions, we tin can apply functional transformations to find the range of a function by graphing functions π‘Ž πœƒ + 𝑏 s i n or π‘Ž πœƒ + 𝑏 c o south for some constants π‘Ž and 𝑏 . Permit us consider only the range of functions of the type π‘Ž πœƒ + 𝑏 s i n , since the range of the latter will exist identical.

We begin with the function southward i n π‘₯ that has the range [ − 1 , 1 ] . Multiplying a part past a positive constant π‘Ž results in a vertical dilation (stretching or contracting) by the calibration factor π‘Ž , which changes the range of the function from [ − one , one ] to [ − π‘Ž , π‘Ž ] . Multiplying a office by a negative constant results in a reflection over the π‘₯ -axis followed by a dilation past the calibration factor | π‘Ž | . In this case, the reflection over the π‘₯ -axis does not alter the range of this office since it is symmetric with respect to the π‘₯ -centrality. So, the vertical dilation past the cistron | π‘Ž | makes the range of the role π‘Ž π‘₯ south i n exist [ − | π‘Ž | , | π‘Ž | ] . We annotation that this expression for the range is true whether π‘Ž > 0 or π‘Ž < 0 .

Next, we know that adding 𝑏 to the role results in a vertical shift (upwardly if 𝑏 > 0 and downwardly if 𝑏 < 0 ) by 𝑏 units. Since the range of π‘Ž π‘₯ southward i n is [ − | π‘Ž | , | π‘Ž | ] , a vertical shift by 𝑏 units would change this range to [ − | π‘Ž | + 𝑏 , | π‘Ž | + 𝑏 ] .

For example, let us consider the range of 2 π‘₯ + i south i northward using the post-obit diagram.

In the diagram, the solid blue graph represents the function due south i n π‘₯ that has the range [ − 1 , 1 ] . Multiplying southward i n π‘₯ past ii changes the range from [ − ane , 1 ] to [ − 2 , 2 ] . Two sided blue arrows in the diagram indicate that the original graph is stretched vertically by a factor of 2 to obtain the graph of 2 π‘₯ s i due north , giving the dashed bend. As nosotros noted earlier, we can see that the range of 2 π‘₯ southward i north is [ − 2 , 2 ] . Adding one to two π‘₯ s i n shifts the range up by one, leading to the new range [ − 1 , 3 ] . The vertical red arrows indicate that the graph of ii π‘₯ due south i due north is shifted upward to obtain the graph of 2 π‘₯ + ane southward i northward . We can notation that the range of this red graph is [ − 1 , three ] , every bit expected.

In the adjacent example, we will determine the range of a sine office from its algebraic expression.

Example 3: Finding the Range of a Given Sine Function

Find the range of the part 𝑓 ( πœƒ ) = eight seven πœƒ s i n .

Answer

Retrieve that the sine function, s i n πœƒ , is periodic and its graph oscillates between − 1 and 1. This tells us that the range of south i n πœƒ is [ − i , 1 ] . Nosotros can utilise this data to decide the range of the given role.

The range is the ready of all possible office values, so we are looking for all possible values of the expression eight vii πœƒ , πœƒ . s i n west h e r e i s a due north y r e a l n u thou b e r

Since πœƒ can accept any value, multiplying πœƒ past 7 does not change the set of values that tin result from this expression. Since s i due north πœƒ can accept whatever values in [ − one , ane ] , nosotros know that s i n 7 πœƒ is likewise limited to the same range.

Multiplying 8 to the expression due south i north 7 πœƒ stretches the graph of the role vertically past a cistron of eight. This transformation changes the range from [ − ane , 1 ] to [ − 8 , 8 ] .

Hence, the range of 𝑓 ( πœƒ ) = 8 7 πœƒ south i n is [ − 8 , 8 ] .

In our next example, we will determine the domain and the range of a cosine function using the same method.

Example 4: Finding Domain and Range of Trigonometric Functions

Consider the part 𝑓 ( π‘₯ ) = iv ( seven π‘₯ + πœ‹ ) + 5 c o southward .

  1. What is the domain of 𝑓 ( π‘₯ ) ?
  2. What is the range of 𝑓 ( π‘₯ ) ?

Answer

Part 1

Permit us notice the domain of 𝑓 ( π‘₯ ) . The domain of a office is the set of all possible input values. We know that the domain of role c o south πœƒ is all existent numbers. This tells us that in that location is no restriction for input values for cosine. In the function 𝑓 ( π‘₯ ) , the expression 7 π‘₯ + πœ‹ is inside the cosine part. Since this function is well divers for any existent number, the domain of 𝑓 ( π‘₯ ) is all real numbers, or ] − ∞ , ∞ [ .

Part 2

Let us consider the range of 𝑓 ( π‘₯ ) . The range is the prepare of all possible part values, and so we need to decide the set of all possible values for the expression 4 ( 7 π‘₯ + πœ‹ ) + 5 , π‘₯ . c o s f o r a n y r e a l n u m b east r

We know that the range of vii π‘₯ + πœ‹ is all real numbers, and then this expression can take any value. Denoting πœƒ = 7 π‘₯ + πœ‹ , nosotros demand to notice the set of possible values of the expression 4 πœƒ + five , πœƒ . c o s f o r a n y r e a l n u m b e r

Let us think through part transformations to obtain the range of this expression. Nosotros know that c o s πœƒ has the range [ − one , 1 ] . Multiplying the role by four results in stretching this range vertically by the factor of 4, leading to the range [ − 4 , 4 ] . Adding 5 to this expression shifts the function upwards past 5, which gives u.s.a. the range [ i , 9 ] .

Alternatively, we tin can find this algebraically by performing the following computations: − 1 ≤ πœƒ ≤ 1 − 4 ≤ 4 πœƒ ≤ 4 i ≤ four πœƒ + 5 ≤ 9 . c o s c o s c o southward

This also leads to the range [ ane , 9 ] .

The range of 𝑓 ( π‘₯ ) is [ 1 , 9 ] .

In our next instance, nosotros volition identify an unknown constant in a trigonometric part when we are given the range of the function.

Example v: Finding the Range of a Trigonometric Function from Its Rule

The range of the role 𝑓 ( πœƒ ) = π‘Ž 3 πœƒ c o s is − five 4 , 5 4 . Find the value of π‘Ž where π‘Ž > 0 .

Reply

Recall that the range of c o s πœƒ is [ − 1 , 1 ] and the domain of c o due south πœƒ is all real numbers. Since both iii πœƒ and πœƒ have the same range, c o s 3 πœƒ has the same range every bit c o due south πœƒ . This tells usa that the gear up of all possible outcomes, that is, the range, of c o due south 3 πœƒ is [ − 1 , 1 ] .

Multiplying a function by a positive constant π‘Ž results in a vertical dilation past the cistron π‘Ž . Since the range of c o s 3 πœƒ is [ − i , 1 ] , applying a vertical dilation to this range makes the range of π‘Ž 3 πœƒ c o s be [ − π‘Ž , π‘Ž ] .

We have shown that the range of 𝑓 ( πœƒ ) = π‘Ž 3 πœƒ c o southward is [ − π‘Ž , π‘Ž ] . We are given that the range of this function is − v 4 , 5 four . This leads to π‘Ž = 5 4 .

And so far, we have considered examples dealing with the domain and the range of functions involving sines and cosines. Let us plough our focus to the tangent function. Think that the tangent part is divers as the ratio of the sine and cosine functions t a n s i north c o south πœƒ = πœƒ πœƒ .

This means that the tangent function will have domain restrictions when the cosine role equals zero. Consider the graph t a n πœƒ over the interval of [ − iii πœ‹ , 3 πœ‹ ] radians.

From the graph, nosotros can meet that the tangent office repeats over πœ‹ radians. Hence, the period of the tangent part is πœ‹ radians, which is dissimilar from the period of the sine and cosine functions. This ways that we tin can find the domain and range of the tangent part past examining its graph over the interval [ 0 , πœ‹ ] . Over this interval, nosotros meet that the tangent office is not defined at πœƒ = πœ‹ ii . Since this function has period πœ‹ , this means that the tangent function is not defined at every πœ‹ starting from the point πœ‹ 2 , as we tin see in the graph. In other words, t a n πœƒ is undefined when πœƒ = πœ‹ two + 𝑛 πœ‹ , 𝑛 ∈ β„€ .

To find the range of t a n πœƒ , nosotros can also consider its graph over the interval [ 0 , πœ‹ ] . Nosotros tin see that the graph of the function over this interval goes upward to the positive infinity and down to the negative infinity. This means that the range of t a n πœƒ is the fix of all real numbers.

Nosotros summarize these results every bit follows.

Definition: Domain and Range of Tangent Function

The domain of t a north πœƒ , in radians, is all real numbers except for πœƒ = πœ‹ 2 + 𝑛 πœ‹ , 𝑛 ∈ β„€ .

The domain of t a n πœƒ , in degrees, is all existent numbers except for πœƒ = nine 0 + 1 8 0 𝑛 , 𝑛 ∈ β„€ . ∘ ∘

The range of t a n πœƒ is all existent numbers, denoted either ] − ∞ , ∞ [ or ℝ .

In our final instance, we will identify the input values where a tangent function is undefined.

Example 6: Finding the Values Where Tangent is Undefined

Find the values of πœƒ in radians such that the office 𝑓 ( πœƒ ) = ( 3 πœƒ ) t a n is undefined.

Answer

Call back that the domain of the tangent part t a due north πœƒ , in radians, excludes values of the class πœƒ = πœ‹ 2 + 𝑛 πœ‹ , 𝑛 ∈ β„€ .

The given part 𝑓 ( πœƒ ) = ( three πœƒ ) t a north includes the tangent function, so we need to find the values of πœƒ such that this function is not defined when the input 3 πœƒ of the tangent gives one of these values. In other words, 𝑓 ( πœƒ ) is not divers when 3 πœƒ = πœ‹ 2 + 𝑛 πœ‹ , 𝑛 ∈ β„€ .

Dividing both sides of the equation to a higher place by iii, the part 𝑓 ( πœƒ ) = ( iii πœƒ ) t a n is undefined when πœƒ is equal to πœ‹ half dozen + 𝑛 πœ‹ 3 , 𝑛 ∈ β„€ .

Allow us finish by recapping a few important points from the explainer.

Key Points

  • If a function 𝑓 ( π‘₯ ) is periodic with menstruation 𝐿 , then we can discover the domain and range of this part past because the graph of this role over the interval [ 0 , 𝐿 ] .
  • The domain of functions due south i n πœƒ and c o s πœƒ is all existent numbers, denoted either ] − ∞ , ∞ [ or ℝ .
    The range of functions s i northward πœƒ and c o s πœƒ is [ − 1 , 1 ] .
  • For whatsoever constants π‘Ž and 𝑏 , the range of the function π‘Ž π‘₯ + 𝑏 due south i north or π‘Ž π‘₯ + 𝑏 c o s is [ − | π‘Ž | + 𝑏 , | π‘Ž | + 𝑏 ] .
  • The domain of t a n πœƒ , in radians, is all existent numbers except for πœƒ = πœ‹ 2 + 𝑛 πœ‹ , 𝑛 ∈ β„€ . The domain of t a n πœƒ , in degrees, is all real numbers except for πœƒ = 9 0 + 1 8 0 𝑛 , 𝑛 ∈ β„€ . ∘ ∘ The range of t a n πœƒ is all real numbers, denoted either ] − ∞ , ∞ [ or ℝ .

Source: https://www.nagwa.com/en/explainers/156129153287/

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