Question Video: Finding the Inverse of a Square Root Function Algebraically Mathematics

Find 𝑓⁻¹(π‘₯) for 𝑓(π‘₯) = √π‘₯ + 3 and state the domain.

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Video Transcript

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.

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