What is the Range of the Given Function

Learning how to find the range of a function can prove to be very important in Algebra and Calculus, because it gives you the capability to assess what values are reached by a part. Or in other words, it allows you to discover the set of all the images via the function

The task of finding what points tin be reached past a part is a very useful one. For example, you lot may have a product function \(q(x)\), which gives yous the amount of output obtained for \(x\) units of input.

We would like to know how many input units are needed to produce \(b\) units of output. So in other words, we need to find \(10\) so that \(q(ten) = b\), which is some other way of asking whether or not \(b\) is in the range of the office \(q(x)\).

In this tutorial we will concentrate more on the mechanics of finding the range. For a more conceptual arroyo to domain and range, y’all can check this tutorial .

The Algebraic Way of Finding the Range of a Function

Aforementioned as for when we learned how to compute the domain, at that place is non i recipe to discover the range, it actually depends on the structure of the part \(f(10)\).

Yet, there is ane algebraic technique that will e’er be used. This is THE way you detect the range. Pay attention:

Say that we need to become the range of a given function \(f(x)\). Then, we will consider a generic real number \(y\) and nosotros will attempt to solve for \(x\) the following equation:

\[f(x) = y\]

Nosotros demand to make up one’s mind for which values of \(y\) the to a higher place equation can be solved for \(ten\). That is it. Of form, that could be hard to do, depending on the structure of the function \(f(ten)\), but its what you need to do.

So this is the algebraic way, the style how to find range of a office without graphing.

Instance ane

Discover the range of the office \(\displaystyle f(10) = \frac{x+1}{x-3}\):

Reply:

We go on using the algebraic way: Let \(y\) be a number and we will solve for \(x\) in the following equation: \(f(10) = y\). The value \(y\) is in the range if \(f(x) = y\) tin can exist solved for \(x\).

In this case we have:

\[\large f(10) = y \Leftrightarrow \frac{x+1}{ten-3} = y\] \[\Rightarrow \,\,\,x+1 = y\left( x-3 \right)\] \[\Rightarrow \,\,\,ten+1 = yx-3y\] \[\Rightarrow \,\,\,ten-yx=-1-3y\] \[\Rightarrow \,\,\,x\left( one-y \right)=-one-3y\] \[\Rightarrow \,\,\,x=\frac{3y+i}{y-1}\]

Therefore, when will \(ten\) exist well defined? Almost for all \(y\), except for when \(y = 1\), considering in that case we have a segmentation past \(0\). Hence, the range of \(f\) in this case is the whole real line, except for one.

If nosotros use interval note, we can write \(Range(f) = (-\infty, i) \cup (one, +\infty)\).

EXAMPLE 2

Observe the range of the function \(f(x) = x^ii – 4x + iii\):

ANSWER:

Again, we go along using the algebraic fashion, then you know the drill: Let \(y\) be a number and we volition solve for \(10\) in the following equation: \(f(x) = y\). The value \(y\) is in the range if \(f(x) = y\) can be solved for \(10\).

In this instance nosotros take:

\[\large f(x) = y \Leftrightarrow 10^ii – 4x + 3 = y\] \[\Rightarrow \,\,\, x^2 – 4x + iii – y = 0 \text{ (This is a quadratic equation in 10)}\] \[\Rightarrow \displaystyle \,\,\, ten = \frac{-b \pm \sqrt{b^two – 4ac}}{2a}\] \[\Rightarrow \displaystyle \,\,\, x = \frac{-(-four) \pm \sqrt{(-4)^2 – 4(1)(3-y)}}{2(ane)}\] \[\Rightarrow \displaystyle \,\,\, ten = \frac{4 \pm \sqrt{16 – iv(3-y)}}{2}\] \[\Rightarrow \displaystyle \,\,\, x = \frac{4 \pm \sqrt{16 – 12 + 4y}}{2}\] \[\Rightarrow \displaystyle \,\,\, ten = \frac{4 \pm \sqrt{4 + 4y}}{2}\] \[\Rightarrow \displaystyle \,\,\, x = 2 \pm \sqrt{1+y}\]

Now, seeing this final expression, when will \(x\) be well defined? Nosotros need to have that the argument of the foursquare root needs to exist not-negative, so we need:

\[one+y \ge 0\]

which means that \(y \ge -1\). If nosotros use interval notation, we can write \(Range(f) = [-1, +\infty)\).

In this example, we could take solved it using the fact that \(f(10) = x^2 – 4x + 3\) is a quadratic function, and its graph is a parabola that opens upward.

The minimum point of this parabola is reached at the vertex. The x-coordinate of the vertex is:

\[\displaystyle x_V = \frac{-b}{2a} =\frac{-(-iv)}{ii(1)} =\frac{4}{ii} = ii\]

Now, the y-coordinate of the vertex is simply found by plugging the value \(x_V = 2\) into the quadratic function:

\[y_V = f(x_V) = 2^2 – 4(2) + 3 = -i\]

Since the minimum value reached by the parabola is \(-1\), we conclude that the range is \([-1, +\infty)\), which is the same conclusion as the one plant algebraically.

The graph of the role \(f(x) = 10^two – 4x + 3\) makes information technology even more than clear:

Nosotros can see that, based on the graph, the minimum is reached at \(x = 2\), which is exactly what was plant to the 10-coordinate of the vertex.

The take a chance of using the graph to find the range is that you could potentially misread the critical points in the graph and give an inaccurate evaluation of the where the function reaches its maximum or minimum.