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Graphs of Inverses of Functions
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In this video, we will learn how to use a graph to find the inverse of a function and analyze the graphs for the inverse of a function.
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To begin, letβs consider what is meant when we say the inverse of a function.
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A function π can be inverted by the function π when π of π₯ is equal to π¦ and π of π¦ is equal to π₯.
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We can think of these two functions as undoing the actions of each other.
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This means that if we were to apply π and then we were to apply π, it would be the same as doing nothing and the result would be our starting value.
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We can mathematically represent this as a function composition as follows.
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Here we have considered a single input value π₯ and a single output value π¦.
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However, if π of π₯ is a one-to-one function, it can be inverted over its entire domain.
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Weβll return to this βone to oneβ constraint later in this video.
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But for now, we should understand that when π of π₯ is a one-to-one function, we can find a uniquely defined inverse function.
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Rather than calling this function π, itβs common to use the following notation.
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We would read this as π inverse.
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We should also note that π is the inverse function of π inverse.
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So the relationship goes both ways.
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Returning to our function composition, we can say that applying both π and π inverse in either order will give us the same value that we started with.
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Before moving forward, letβs highlight one quick point about notation.
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We should be very careful not to mistake this minus one for an exponent.
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As a very quick example, if we had some function π of π₯ equals two π₯ minus three, π inverse would not be two π₯ minus three all raised to the power of minus one.
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This video will not go into any detail on how to algebraically find the inverse of a function.
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But itβs enough to say this common misconception should be avoided.
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Okay, now that weβve recapped the inverse of a function, letβs see how it relates to graphs.
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Consider some invertable function π.
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When graphing functions, we usually consider the π₯-axis as the domain of the function, which we can think of as an input.
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We consider the π¦-axis as the range, which we can think of as an output.
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Letβs say we were to input some value π into our function which gave an output of π.
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More formally, we might denote this as π of π equals π.
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If we were to plot the graph of our function π of π₯, we would say that the point with coordinates π, π lies on the line or curve.
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Now letβs consider the function π inverse.
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As we saw earlier, π inverse undoes the action of π.
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This means that if we were to input the value of π into π inverse, the resulting output must be π.
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We can follow a similar line of reasoning to conclude that the point with coordinates π, π must lie on the graph of π inverse.
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Letβs consider what we have just found.
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If we find a coordinate π, π which lies on the graph of π of π₯, we can reverse the ordered pair to find a corresponding coordinate π, π which lies on the graph of π inverse.
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We can also think of this as a swap in the π₯- and π¦-coordinate.
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Since we followed a general method to get here for any given point on the graph of π or indeed of π inverse, this relationship must be true.
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Another interesting point, since this relationship is true for all values, we might be able to see that the domain of π is the same as the range as π inverse and the range of π is the same as the domain of π inverse.
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Letβs now pick some values and see what all of this looks like on a graph.
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Consider the function π of π₯ equals two π₯ minus six.
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If we wanted to pick some values to plot in the top-right quadrant for π, we could use the coordinates three, zero; four, two; and five, four.
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Now remember, to find points on the graph of π inverse, we can simply swap our ordered pairs.
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This means that the points zero, three; two, four; and four, five all lie on the graph of π inverse.
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At this point, we might begin to notice a pattern or perhaps even a link to function transformations.
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The change that we have described with the ordered pairs switches the π₯-coordinate for the π¦-coordinate and vice versa.
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This change actually corresponds to a reflection in the line π¦ equals π₯.
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Looking at the graph, we can see that all three of the points weβve plotted display this rule.
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Another thing we might notice is that if our function π intersects the line of reflection π¦ equals π₯ at any point, π inverse will also pass through this point.
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On our graph, this happens at the point six, six.
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Joining up all of our points, we should be able to see our reflection more clearly.
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Again, note that this video will focus on graphs of inverse functions rather than algebraic manipulation.
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The important rule that we have just found relating to this is that inverse functions have graphs which are reflections of each other in the line π¦ equals π₯ and the coordinates of the points on these lines or curves are reversed ordered pairs.
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Let us now take a look at an example of a question.
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The following is the graph of π of π₯ equals two π₯ minus one.
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Which is the graph of the inverse function π inverse of π₯?
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And just to be clear for this question, π of π₯ is the blue line, and our options for π inverse of π₯ are shown here.
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When trying to find graphs of inverse functions, we should remember the following important rule.
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Inverse functions have graphs which are reflections in the line π¦ equals π₯ and have corresponding coordinates which are reversed ordered pairs.
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One approach to solving this problem would be to draw the line π¦ equals π₯ onto our diagram.
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We would then reflect the graph of π of π₯ using the line π¦ equals π₯ as our line of symmetry.
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Note here that the question has tried to throw in some confusing factors.
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For the graph given by the question, the π₯-axis goes from negative four to four and the π¦-axis goes from a negative six to six.
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This is not the same for the axes in all of our options.
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In spite of this, we can still begin to eliminate some options.
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We can first observe that the graph of π inverse of π₯ has a positive gradient.
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Looking at options (b) and (c), we can see both of these have a negative gradient.
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They can therefore be eliminated as they cannot be the graph of π inverse.
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Next, we can observe that the graph of π inverse appears to have a positive π¦-intercept.
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Of our remaining two options, option (a) does have a positive π¦-intercept, whereas option (d) does not.
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This means that we can eliminate option (d).
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This means that the answer to our question must be option (a).
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We might also want to consider that there are other ways that we could have solved this problem.
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In particular, we can use the fact that corresponding coordinates are reversed ordered pairs.
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Imagine we had not reflected π of π₯ in the line π¦ equals π₯.
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Instead, we could eliminate the incorrect options by cleverly picking a point on the graph of π of π₯.
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Letβs pick the π¦-intercept which is the point negative one, zero.
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Reversing this ordered pair gives us the coordinates of a corresponding point which we know must lie on the line of π inverse of π₯.
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That coordinate is zero, negative one.
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We can then use this information by looking at all of the options for the graph of π inverse of π₯ and seeing which of these graphs appears to pass through the point zero, negative one.
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Of course, we already know the answer.
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The graph of option (a) does pass through the point zero, negative one.
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The graphs of option (b), (c), and (d) do not.
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Again, we have confirmed that the graph of π inverse of π₯ is option (a).
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Letβs now take a look at another example.
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Shown is the graph of π of π₯ equals five π₯ cubed plus six.
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Find the intersection of the inverse function π inverse of π₯ with the π₯-axis.
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To begin, we might remember the following rule.
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Inverse functions have graphs which are reflections of each other in the line π¦ equals π₯.
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One approach to solving this problem might be to draw the line π¦ equals π₯.
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We could then reflect the graph of π of π₯ using π¦ equals π₯ as our line of symmetry and then find the intersection of π inverse of π₯ with the π₯-axis.
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This alternate approach relies on the fact that coordinates of points which lie on the graphs of inverse functions are reversed ordered pairs.
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In other words, if we swap the π₯- and π¦-coordinates for a point on the graph of π, we get the coordinates of a corresponding point which we know is on the graph of π inverse.
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Also note that this relationship works both ways.
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Okay, so weβre looking for the intersection of the graph of π inverse with the π₯-axis.
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This intersection point will have a π¦-coordinate of zero.
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Letβs say this intersection point on the graph of π inverse has the coordinates π, zero.
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Remember that we can swap this ordered pair to get the corresponding point on the graph of π.
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The corresponding point on the graph of π will therefore have the coordinates of zero, π.
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In other words, it will be the π¦-axis intercept on the graph of π.
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Okay, so our method will then be to find the π¦-axis intercept on the graph of π and to swap the ordered pair to get the π₯-axis intercept on the graph of π inverse.
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Looking at the graph of π given in the question, we can clearly see that the π¦-axis intercept has coordinates zero, six.
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Swapping this ordered pair, we conclude that the corresponding point on the graph of π inverse will have the coordinates six, zero.
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With this, we have answered our question.
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The graph of π inverse of π₯ intersects the π₯-axis at the point six, zero.
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Okay, earlier in this video, we briefly mentioned one-to-one functions and how they related to inverse functions.
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Letβs explore this in more detail now.
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We said that if π is a one-to-one function, it has an inverse function.
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It is also true that if π is not a one-to-one function, it does not have an inverse.
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Let us explore why this is the case using the following graph.
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This function π of π₯ is not a one-to-one function.
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One of the ways that we can test if something is a one-to-one function is using the horizontal line test.
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This says that if we can draw a horizontal line that intersects the graph at more than one point, the graph does not represent a one-to-one function, and we say it fails the test.
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For our graph of π of π₯, we can clearly see that a horizontal line can be drawn such that it intersects the graph at more than one point.
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This means that π of π₯ fails the test, and we confirm it is not a one-to-one function.
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Okay, we know that reflecting a graph in the line π¦ equals π₯ gives us the graph of its inverse.
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Let us see what happens in this case.
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Now, we could choose to do this on the same set of axes, but letβs do this on a different set for clarity.
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We will tentatively call this new graph π inverse of π₯.
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But as weβll see soon, this is not the case.
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Looking at this new graph, we might notice an important detail.
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Remember that in order for some relation to be called a function, each input must correspond to only one output.
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One way to test for this is using the vertical line test.
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Weβre gonna edit this statement below to now represent the vertical line test.
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This test says that if we can draw a vertical line that intersects the graph at more than one point, the graph does not represent a function.
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Earlier in this video, you may recall that we said the domain and range of some function can be said to swap for its inverse function.
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If we could draw a horizontal line that intersected our original graph of π of π₯ at more than one point, it might now be obvious that we can draw a vertical line that intersects our new graph at more than one point.
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We can also clearly see this on our diagram.
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This means that what we have tentatively called π inverse of π₯ is not a function at all.
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Since this is not a function, we say that our original function π of π₯ does not have an inverse.
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Another way to think about this is as follows.
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Consider our original function π.
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There are two values in its domain, π one and π two, which correspond to the same value in its range, π one.
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This means that for the relation which we supposed was the inverse of π, which we now know is not a function, the value of π one, which is now in its domain, must correspond to two values, π one and π two, in its range.
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Consider what would happen if we try to use this relation to undo the action of our original function π for an output of π one.
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Our supposed inverse cannot tell us for certain whether our original input was π one or π two.
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For this reason, we say that if something is not a one-to-one function, it does not have an inverse.
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Letβs take a look at an example to illustrate this.
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Determine which of the following functions does not have an inverse.
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To answer this question, weβre going to be using the fact that if a function is not a one-to-one function, it does not have an inverse.
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The way that we can test for one-to-one functions is using the horizontal line test.
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This tells us that if we can draw a horizontal line that intersects the graph at more than one point, it does not represent a one-to-one function.
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And we say the graph fails the test.
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After carefully examining our options, it should become clear that there is only one graph that we could draw a horizontal line on so that it intersects the graph at more than one point.
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This is the function π of π₯.
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We have just found that π of π₯ fails the horizontal line test.
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This means that itβs therefore not a one-to-one function, and so it does not have an inverse function.
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The answer to our question is therefore option (b).
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π of π₯ does not have an inverse function, but the other three options do since they passed the horizontal line test.
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Letβs now take a look at one final example.
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By sketching graphs of the following functions, which is the inverse of itself?
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This question has directed us to sketch these four graphs, and so this should be our first step.
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At this point, you should be familiar with sketching graphs.
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And in order to avoid going into unnecessary detail, this video will simply provide them.
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You may, however, wish to verify these graphs yourself by drawing up a small table of values or by using graphing software.
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First, we have the graph of one over π₯.
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Next, we have the graph of π₯ squared, forming the familiar shape of a parabola.
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We then have the graph of π₯ cubed and, finally, the graph of one over π₯ squared.
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In order to move forward with this question, we recall the following rule.
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Inverse functions have graphs which are reflections of each other in the line π¦ equals π₯.
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This means that a graph which is the inverse of itself has symmetry about the line π¦ equals π₯.
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What this means is that if a function is the inverse of itself when reflected over the line π¦ equals π₯, the resulting function will be the same as the original function.
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You may already be able to spot the answer, but letβs reflect all of our functions in the line π¦ equals π₯ now.
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We first reflect one over π₯, then π₯ squared, followed by π₯ cubed, and finally one over π₯ squared.
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After doing so, it should be clear to us that the reflection of the graph one over π₯ is the same.
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Itβs also the graph of one over π₯.
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The reflections of all three of the other options are different.
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This means that the inverse functions are different from the original function.
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Hence, π₯ squared, π₯ cubed, and one over π₯ squared are not inverses of themselves.
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One small side note, for options (b) and (d), the reflections are not actually functions at all.
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These graphs would fail the vertical line test.
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Since the graphs are not functions at all, we say that π₯ squared and one over π₯ squared do not have inverses.
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Okay, back to our question, it should be clear that option (a) is the only graph for which the original and the reflection are the same.
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This means that from our options one over π₯ is the only function which is the inverse of itself.
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To finish off this video, letβs go through some key points.
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If π is a one-to-one function, it has an inverse function which can be represented using the following notation.
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We would say this as π inverse.
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If π is not a one-to-one function, it does not have an inverse function.
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We can think of inverse functions as undoing the action of each other.
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This can be represented as a function composition like so.
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Inverse functions have graphs which are reflections of each other in the line π¦ equals π₯.
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The coordinates of points which lie on the graphs of inverse functions can be thought of as reversed ordered pairs.
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We saw that this can be interpreted as the domain and the range of a function switching for its inverse.
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Finally, some functions are their own inverse.
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The graphs of these functions have reflectional symmetry in the line π¦ equals π₯.