Lesson Video: Domain and Range of a Function | Nagwa Lesson Video: Domain and Range of a Function | Nagwa

Lesson Video: Domain and Range of a Function Mathematics • Second Year of Secondary School

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In this video, we will learn how to identify the domain and range of functions from their equations.

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Video Transcript

In this video, we will learn how to identify the domain and range of functions from their equations.

We begin by recalling that a function 𝑓 assigns to members of one set, for example, π‘₯, values in a second set π‘Œ. We think of π‘₯ as the source of inputs to the function and π‘Œ as the target of its outputs. For every input π‘₯ that exists in set 𝑋, we have an output 𝑓 of π‘₯ that exists in set π‘Œ. This leads us to our definitions for the domain and range. The domain of 𝑓 is the set of all inputs π‘₯ that exist in set 𝑋 for which 𝑓 of π‘₯ is defined. And the range of 𝑓 is the set of all outputs 𝑓 of π‘₯ as π‘₯ varies over the domain.

In our first example, we will identify the ordered pairs of a function given the domain, range, and the equation of the function.

𝑋 and π‘Œ are two sets of numbers where set 𝑋 contains the values 10, one, two, and eight and set π‘Œ contains the values 12, seven, 60, six, 48, and four. The function 𝑓 of π‘₯ is equal to six π‘₯, where 𝑓 maps elements of 𝑋 onto elements of π‘Œ. Find the ordered pairs that satisfy the function and its range.

One way of answering this question is to consider a mapping diagram that represents the function 𝑓. There are four elements in set 𝑋. They are the numbers 10, one, two, and eight. Set π‘Œ contains six elements, the values 12, seven, 60, six, 48, and four. We are told that the function 𝑓 of π‘₯ is equal to six π‘₯. We recall that the domain of 𝑓 is the set of inputs for which 𝑓 of π‘₯ is defined. And the range of π‘₯ is the set of corresponding outputs.

Our first value in set 𝑋 is 10. And since 𝑓 of π‘₯ is equal to six π‘₯, 𝑓 of 10 is equal to six multiplied by 10. This is equal to 60 and means that an input of 10 gives an output of 60. The first ordered pair that satisfies the function is 10, 60. 𝑓 of one is equal to six multiplied by one, which equals six. This gives us a second ordered pair one, six. Repeating this for the third value in set 𝑋, we have 𝑓 of two is equal to 12, giving us a third ordered pair two, 12. Finally, 𝑓 of eight is equal to 48. And we have a fourth ordered pair eight, 48. The four ordered pairs that satisfy the function are 10, 60; one, six; two, 12; and eight, 48.

We are also asked to give the range of the function. As already mentioned, this is the set of all outputs of 𝑓 of π‘₯. The range of the function is therefore the set of the four values 60, six, 12, and 48. And we have now answered the two parts of this question.

In our next question, we will identify the range of a function represented by a Cartesian diagram.

The figure below shows the graph of a function 𝑓. What is the range of the function?

We can see from the figure that the function 𝑓 contains four ordered pairs. They are one, negative two; two, negative three; three, zero; and four, negative three. We are asked to find the range of this function.

We begin by recalling that the domain of a function is the set of all input or π‘₯-values for which the function 𝑓 of π‘₯ is defined. These are the π‘₯-values of our ordered pairs. The numbers one, two, three, and four. The range of a function is the set of all output values. These are the 𝑦-coordinates in our ordered pairs. We have the values negative two, negative three, zero, and negative three once again.

Despite the fact that this value appears twice, we only need to include it once when writing the range. The range of the function shown in the figure is the set of values negative two, negative three, and zero. And writing these in ascending order, we have negative three, negative two, zero.

We could also find these values directly from the figure by drawing horizontal lines through all the points on the graph. These three lines intersect the 𝑦-axis at negative three, negative two, and zero.

In our next two examples, we will identify the domain and range of a function given its graph.

Determine the domain of the function represented in the graph below.

We begin by recalling that the domain of a function 𝑓 is the set of all π‘₯-values or inputs for which 𝑓 of π‘₯ is defined. The solid dot at the far left of our curve lies at the point with coordinates four, one. This means that when π‘₯ equals four, 𝑦 is equal to one. And using function notation, 𝑓 of four equals one.

The arrow at the other end of our curve tells us that the function is defined for all real numbers to the right of this point. And we can therefore conclude that the function is defined for all values of π‘₯ greater than or equal to four. Had the dot at the point four, one been hollow, then 𝑓 of π‘₯ would’ve been defined on the strict inequality π‘₯ is greater than four. This inequality π‘₯ is greater than or equal to four is the domain of the function. And we can also write this using interval notation. The domain of the function represented in the graph is the left-closed, right-open interval from four to ∞.

Whilst we could also find the range of the function from the graph, we will do this in the next example.

Determine the range of the function represented by the graph below.

We begin this question by recalling that the range of a function 𝑓 is the set of all its outputs or 𝑦-values. We need to find all the 𝑦-values that are represented by the curve. Our graph is in the shape of a parabola. And it appears that the function is quadratic. It has a maximum at the point with coordinates negative seven, one. This means that 𝑓 of negative seven is equal to one, and the maximum value of the function is one. There is no π‘₯-value such that 𝑓 of π‘₯ is greater than one.

Looking at the graph, it does appear that the parabola takes on all values less than or equal to one. And we can therefore conclude that the range of the function is the set of real 𝑦-values such that 𝑦 is less than or equal to one. This can also be written using interval notation as the left-open, right-closed interval from negative ∞ to one.

In our final two examples, we will identify the domain and range of a function given its equation.

Determine the domain of the function 𝑓 of π‘₯ which is equal to the square root of the absolute value of π‘₯ minus 33.

We begin by recalling that the domain of a function is the set of inputs for which 𝑓 of π‘₯ is defined. We know that the square root is only valid for nonnegative numbers. This means that the absolute value of π‘₯ minus 33 must be greater than or equal to zero. Adding 33 to both sides of our inequality, we have the absolute value of π‘₯ is greater than or equal to 33. Recalling the graph of the equation 𝑦 is equal to the absolute value of π‘₯ and then drawing the horizontal line 𝑦 is equal to 33, we see that the absolute value of π‘₯ is greater than or equal to 33 when π‘₯ is less than or equal to negative 33 or π‘₯ is greater than or equal to 33. These inequalities can be rewritten as the union of two intervals: the left-open, right-closed interval from negative ∞ to negative 33 and the left-closed, right-open interval from 33 to ∞.

Whilst this is a perfectly valid answer for the domain of the function, this is not an interval. We can, however, write it as the complement of one. The domain of the function is the set of real values minus the set on the open interval from negative 33 to 33.

We could once again also find the range of this function. However, we’ll do this in one final example.

If 𝑓 maps elements on the closed interval from two to 21 to the set of real numbers, where 𝑓 of π‘₯ is equal to three π‘₯ minus 10, find the range of 𝑓.

In this question, we are given a linear function 𝑓 of π‘₯, which is equal to three π‘₯ minus 10. We know that the graph of 𝑦 is equal to three π‘₯ minus 10 is a straight line as shown. We know that the domain of a function is the set of input or π‘₯-values. And in this question, we’re told that these lie on the closed interval from two to 21. The function 𝑓 of π‘₯ is therefore defined for all values between the two points shown.

We can calculate the corresponding 𝑓 of π‘₯ values by substituting π‘₯ equals two and π‘₯ equals 21 into the function. When π‘₯ is equal to two, 𝑓 of π‘₯ is equal to three multiplied by two minus 10. This is equal to negative four. Likewise, when π‘₯ is equal to 21, 𝑓 of π‘₯ is equal to three multiplied by 21 minus 10. This is equal to 53. Since the range of a function 𝑓 is the set of outputs or 𝑦-values, we can conclude that 𝑓 of π‘₯ or 𝑦 is greater than or equal to negative four and less than or equal to 53. Using interval notation, the range of the function 𝑓 on the given domain is the closed interval from negative four to 53.

We will now summarize the key points from this video. In this video, we introduced the definitions for the domain and range of a function. The domain of a function is the set of inputs or π‘₯-values for which the function is defined. And the range is the set of corresponding outputs or 𝑦-values. We considered examples where we identified the domain and range of a function given a mapping, graph, or the equation of the function.

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