Lesson Video: Inverse of a Function | Nagwa Lesson Video: Inverse of a Function | Nagwa

Lesson Video: Inverse of a Function Mathematics • Second Year of Secondary School

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In this video, we will learn how to find the inverse of a function by changing the subject of the formula.

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Video Transcript

In this video, we will learn how to find the inverse of a function by changing the subject of the formula.

In order to consider the inverse of a function, we must first be familiar with functions themselves. We should recall that a function, which we might call 𝑓, takes an input value π‘₯ and maps it to an output value 𝑓 of π‘₯. It’s defined by an equation showing how the output is calculated from the input. For example, the function 𝑓 of π‘₯ equals two π‘₯ minus three takes a value of π‘₯. It then doubles it and subtracts three to give an output value. If we were to input the value four to this function, we would find that 𝑓 of four is equal to two multiplied by four minus three, that’s eight minus three, which is equal to five. So the input four leads to the output five.

We can also visualize this using a diagram to show the steps involved in mapping an input value to an output value. We multiply four by two to give eight and then subtract three to give five. If we were to plot the graph of 𝑦 equals 𝑓 of π‘₯, we would, in this instance, get a straight line representing the mapping all possible input values of π‘₯ to the output values 𝑓 of π‘₯. We could then use this graph to work out the output value for a given input value. For example, for an input value of two, we would go up from two on the π‘₯-axis until we meet the line. We then go across from the line to the 𝑦-axis, and we find that the output value for an input value of two is one.

We can also use this graph to illustrate what we mean by the domain and range of a function. The domain of a function is the set of all possible input values, which corresponds to all values on the π‘₯-axis for which the graph can be plotted. In this case, the domain of the function is the complete set of real numbers. The range of a function is the set of all outputs, which are all the 𝑦-values on the graph. Because this line continues infinitely in both directions, the range of this function is also the complete set of real numbers.

Now that we’ve recapped the functions themselves, let’s think about their inverses. Suppose we had an output value 𝑦, in this case 𝑦 equals three, and we wanted to find the input value corresponding to it. Well, we could do this from the graph. We could go across from the 𝑦-value of three to the line and then down to the π‘₯-axis to find the corresponding input value, which it happens is also three. But it would be nice to be able to do this without needing to draw the graph. We could substitute this value into our function as the output value and then solve for π‘₯. Doing so, we find that an output value of three is indeed associated with an input value also of three.

But if we wanted to find the input value for many output values, it would be more efficient to find a general equation we can use rather than substituting each value and then solving the equation each time. To do this, we could rearrange the general equation 𝑦 equals 𝑓 of π‘₯ to make π‘₯ the subject. Starting with the equation 𝑦 equals two π‘₯ minus three then, we can add three to both sides to give 𝑦 plus three is equal to two π‘₯ and then divide both sides of the equation by two to give 𝑦 plus three over two equals π‘₯. And we’ve rearranged the formula to make π‘₯ the subject. As before, if we substitute the output or 𝑦-value of three, we get π‘₯ equals three plus three over two, which is six over two, which is equal to three.

But we can use the same equation to find the input value for any output value. For example, when 𝑦 is equal to zero, π‘₯ is zero plus three over two, which is 1.5. The process we’ve just gone through demonstrates how to find the inverse of a function. Informally, the inverse of a function is a function that undoes the original function. If a function 𝑓 applied to an input π‘₯ gives the output 𝑦, then the inverse of 𝑓, which we represent as 𝑓 and then a superscript negative one, takes the input value 𝑦 and gives the output value π‘₯. So we get back to where we started. We’ve seen that in this example, if we’re given a value of π‘₯, we can find the value of 𝑦, which is equal to 𝑓 of π‘₯, using the equation 𝑦 equals two π‘₯ minus three. And if we’re given the value of 𝑦, or 𝑓 of π‘₯, we can find the value of π‘₯ using the equation π‘₯ equals 𝑦 plus three over two.

We want to write the inverse function as a function of π‘₯ though, not a function of 𝑦. We therefore swab the position of π‘₯ and 𝑦 around, and we define 𝑓 inverse of π‘₯ to be equal to 𝑦. We have that the inverse function is 𝑓 inverse of π‘₯ is equal to π‘₯ plus three over two. We can also consider the link between the graphs of a function and its inverse. By swapping the roles of π‘₯ and 𝑦 in the function around, this means that the graph is flipped so that the 𝑦-axis for the original function is in the position of the π‘₯-axis for the inverse and vice versa.

The geometric transformation taking place here is a reflection in the line 𝑦 equals π‘₯. We can see this if we add the line 𝑦 equals 𝑓 inverse of π‘₯ to our sketch. Because we swap the π‘₯- and 𝑦-axes, it also follows that the domain of the function 𝑓 of π‘₯ becomes the range of the function 𝑓 inverse of π‘₯. And the range of the function 𝑓 of π‘₯ is the domain of its inverse function.

We can formalize the general process of finding an inverse function as follows. First, we let 𝑦 equal the function 𝑓 of π‘₯. Then, we rearrange to make π‘₯ the subject. We then swap π‘₯ and 𝑦 around so that instead of having π‘₯ is equal to some function of 𝑦, we have 𝑦 is equal to some function of π‘₯. Finally, we define this new function of π‘₯ to be 𝑓 inverse of π‘₯. Let’s now consider some examples in which we’ll find the inverse of a function algebraically, beginning with a linear function.

Find 𝑓 inverse of π‘₯ for 𝑓 of π‘₯ equals a half π‘₯ plus three.

We’ve been given a function 𝑓 of π‘₯ and asked to find an expression for its inverse. We begin by letting 𝑦 equal 𝑓 of π‘₯. We then want to rearrange this equation to make π‘₯ the subject. The first step is to subtract three from each side, giving 𝑦 minus three equals a half π‘₯. We can then multiply both sides of the equation by two, giving a two multiplied by 𝑦 minus three is equal to π‘₯. And distributing the parentheses on the left-hand side, we have two 𝑦 minus six is equal to π‘₯. Now, this is π‘₯ as a function of 𝑦, but we want to write the inverse function as a function of π‘₯. We therefore swap the positions of π‘₯ and 𝑦 around. So where we had 𝑦, we swap this for π‘₯, and where we had π‘₯, we change this for 𝑦. So we now have 𝑦 is equal to two π‘₯ minus six.

The final step is to define this new function of π‘₯ to be 𝑓 inverse of π‘₯. So we have that 𝑓 inverse of π‘₯ is equal to two π‘₯ minus six. This is the function that maps all of the old output values back to their original inputs. We can see this if we take an input value, for example, four. When we apply the function 𝑓 of π‘₯ to the value four, we get the output value five. And when we apply this inverse function that we’ve just found to the output value five, we have that 𝑓 inverse of five is equal to two multiplied by five minus six, that’s 10 minus six, which gives the original input value of four.

It’s also worth pointing out that the two steps of rearranging to make π‘₯ the subject and then swapping π‘₯ and 𝑦 around can be done in either order. If you prefer, you can swap π‘₯ and 𝑦 at the beginning and then rearrange to make 𝑦 the subject. We’ll get the same answer, which is that 𝑓 inverse of π‘₯ is equal to two π‘₯ minus six.

Let’s now consider a slightly more complicated example in which the function we’re finding the inverse of is a square root function. We’ll also find the domain of this inverse function.

Find 𝑓 inverse of π‘₯ for 𝑓 of π‘₯ equals the square root of π‘₯ plus three and state the domain.

To find the inverse function, we’ll begin by letting 𝑦 equal 𝑓 of π‘₯. So we have 𝑦 is equal to the square root of π‘₯ plus three. We then rearrange to make π‘₯ the subject. We begin by subtracting three from each side to give 𝑦 minus three equals the square root of π‘₯. We then square both sides of the equation or raise each side to the power of two to give 𝑦 minus three squared is equal to π‘₯. We could distribute the parentheses, but there’s no real need to. We now have π‘₯ as a function of 𝑦, but we want the inverse function to be a function of π‘₯. So we swap π‘₯ and 𝑦 around to give π‘₯ minus three squared is equal to 𝑦. Finally, we define the inverse function 𝑓 inverse of π‘₯ to be equal to the expression for 𝑦. So we have that 𝑓 inverse of π‘₯ is equal to π‘₯ minus three all squared.

We’re also asked to find the domain of the function 𝑓 inverse. We recall that the set of all input values to the function 𝑓 inverse is the same as the set of all output values of the function 𝑓. Or, in other words, the domain of the function 𝑓 inverse is the same as the range of the original function 𝑓. So let’s consider the function 𝑓 of π‘₯, and we want to determine its range or set of all output values. We know that the square root function produces nonnegative values. The minimum value it can produce is zero when π‘₯ itself is equal to zero. And there’s no limit to the maximum value it can produce. If we take the square root of a really big positive number, we get another really big positive number.

We then add three to this value. This means that the minimum value of 𝑓 of π‘₯ will be zero plus three, which is three. And as before, there is no maximum value. The range of the function 𝑓 of π‘₯ is therefore all values greater than or equal to three. This set of values is then the same as the domain for the function 𝑓 inverse. But as these are input values for 𝑓 inverse, they are values of π‘₯. We can state then that the domain of 𝑓 inverse of π‘₯ is π‘₯ is greater than or equal to three. So we’ve completed the problem. 𝑓 inverse of π‘₯ is equal to π‘₯ minus three squared, and the domain is π‘₯ greater than or equal to three.

We’ve now seen two examples of how we can find the inverse of a function algebraically. But it isn’t always possible to find the inverse of a function over its entire domain. Let’s consider the function 𝑓 of π‘₯ equals two π‘₯ squared plus three. The graph of 𝑦 equals 𝑓 of π‘₯ looks like this. It is a positive parabola with a 𝑦-intercept of three. If we have an input value π‘₯, we can find an output value 𝑦 by going up to the graph and then across to the 𝑦-axis. But what if we want to go the other way?

Suppose we have the output value five, and we want to find the input value that maps to this. If we go across from the 𝑦-value of five to the graph, we can see that there are, in fact, two points that have a 𝑦-coordinate of five. This means that there’re two input values, which in fact are positive and negative one, associated with this output value. This means that the inverse function isn’t well defined. If we have an output value of five, how do we know whether this has come from an input value of one or negative one?

This leads to an important condition for the inverse of a function to exist. The inverse of a function only exists over its entire domain if the function is one-to-one. This means that every input to the function produces a unique output so that when we go the other way using the inverse function, every input to that function also produces a unique output. If a function isn’t one-to-one over its entire domain, we can restrict the domain to a set of values over which it is one-to-one and find the inverse only for that restricted domain. For example, for this quadratic function, we could restrict the domain to be only π‘₯-values greater than or equal to zero. Or, indeed, we could use π‘₯-values less than or equal to zero. And then the function would be one-to-one over this restricted domain, and so it would be possible to find its inverse.

Let’s consider now an example of identifying whether a function is invertible over its entire domain.

Which of the following functions does not have an inverse over its whole domain? (a) 𝑓 of π‘₯ equals two π‘₯. (b) 𝑓 of π‘₯ equals two to the power of π‘₯. (c) 𝑓 of π‘₯ equals one over π‘₯. Or (d) 𝑓 of π‘₯ equals π‘₯ squared.

To answer this question, we need to consider the conditions under which a function has an inverse. A function only has an inverse over its entire domain if it is a one-to-one function. This means that every input to the function produces a unique output so that when we go the other way using the inverse function, every input to this function also produces a unique output. We can work out which of these functions isn’t one-to-one by sketching graphs of each function. For option (a), the graph of 𝑦 equals 𝑓 of π‘₯ or 𝑦 equals two π‘₯ is a straight line graph through the origin. This is clearly a one-to-one function over its entire domain. Every input has a unique output. And so going the other way, the same will be true.

More generally though, we can determine whether a function is one-to-one by drawing horizontal and vertical lines on its graph. If each horizontal and each vertical line intersects the graph a maximum of one time, then the function is one-to-one. The graph of 𝑦 equals two to the power of π‘₯ is an exponential graph passing through the point zero, one with a horizontal asymptote along the π‘₯-axis. This is also a one-to-one function over its entire domain because every vertical line intersects the graph once and every horizontal line intersects the graph a maximum of one time.

For the graph of 𝑦 equals one over π‘₯, there’re two branches, one in the first quadrant, where π‘₯ and 𝑦 are both positive, and one in the third quadrant, where π‘₯ and 𝑦 are both negative. We can see, though, that the function is one-to-one over its entire domain. That just leaves option (d) then. The graph of 𝑦 equals π‘₯ squared is a positive parabola with its minimum point at the origin. We can see that if we draw horizontal lines across the graph, some of these lines will intersect the graph in more than one place. This means that for certain output values, there are two possible input values, and so the function 𝑓 of π‘₯ isn’t one-to-one. This means we can’t find the inverse of 𝑓 of π‘₯ over its entire domain because we don’t know which of those input values the output value should be mapped to by the inverse function. Our answer is that the function which does not have an inverse over its whole domain is 𝑓 of π‘₯ equals π‘₯ squared.

Let’s now summarize the key points from this video. Informally, we can think of the inverse of a function as the function which undoes the original function. More formally, for a value π‘₯ in the domain of the function 𝑓, if 𝑓 of π‘₯ is equal to 𝑦, then 𝑓 inverse of 𝑦 is equal to π‘₯. Values in the domain of 𝑓 of π‘₯ form the range of 𝑓 inverse of π‘₯ and vice versa. Values in the range of 𝑓 of π‘₯ form the domain of 𝑓 inverse of π‘₯. We saw that the inverse of a function only exists if the function is one-to-one so that each input value is mapped to a unique output value. And then going the other way, the inverse function maps each output value back to a unique input value.

We also saw that to find the inverse of a function 𝑦 equals 𝑓 of π‘₯, we need to rearrange the formula to make π‘₯ the subject. And then we swap π‘₯ and 𝑦 around. These two steps can be performed in either order. We can also swap π‘₯ and 𝑦 first and then rearrange to make the new 𝑦 the subject. Finally, we also saw that due to this swapping of π‘₯ and 𝑦, the graphs of a function and its inverse are reflections of one another in the line 𝑦 equals π‘₯.

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