The graphs of y = sin⁻¹ x, y = cos⁻¹ x, and y = tan⁻¹ x are shown...

Question

The graphs of y = sin⁻¹ x, y = cos⁻¹ x, and y = tan⁻¹ x are shown in Table 2.8. In Exercises 97–106, use transformations (vertical shifts, horizontal shifts, reflections, stretching, or shrinking) of these graphs to graph each function. Then use interval notation to give the function's domain and range.

h(x) = −2 tan⁻¹ x

Accepted Answer

Welcome back. I am so glad you're here. We're asked to perform transformation on the given graph of Y equals the inverse tangent of X to graph. The following function then identify the domain and range express the domain and range in interval notation. The function that we are being asked to graph is F of X equals negative five inverse tangent of X. And then we are given a graph of Y equals the inverse tangent of X. We have a vertical Y axis, a horizontal X axis. They come together at the origin in the middle. And then the range for our graph is going to be from negative pi divided by two to pi divided by two, not inclusive of either of those because there are horizontal asymptotes at Y equals negative pi divided by two and pi divided by two. The domain for our graph is from negative infinity to positive infinity. And then our graph of Y equals the tangent or the inverse tangent of X is heading toward negative infinity for the X values along that horizontal asom tote Y equals negative pi divided by two. And then it curves up, goes through the origin and then is heading toward positive infinity for the X values along that horizontal Asymptote of Y equals pi divided by two. All right. So we are asked to make transformations from that graph to our graph of F of X equals negative five inverse tangent of X. The big difference here is that our inverse tangent of X is being multiplied by negative five. And that actually indicates two transformations that are taking place first, this negative that the inverse tangent of X is being multiplied by the negative is not being multiplied directly to our X. And when that's happening, that indicates that it's going to be reflected over the X axis. Now, the second transformation that's taking place here is the multiplication of that five. So looking at just the five, that is a number larger than one. And because we're multiplying that again, not directly to the X and it's a number larger than one, it indicates vertical stretching. So those are the two transformations that are going to take place. We can get a pretty good idea of what it looks like with reflection over the X axis and vertical stretching. But we can also be more precise in the transformations that are taking place. So let's look a little bit closer at Y equals the inverse tangent of X and identify a few of our points. We know it's going through the origin at 00. And with both of the two transformations we've identified that's not going to change, it's still going to pass through the origin. But we can identify a couple more points. If we look at the X value of negative one, then we have a Y value of a negative pi divided by four. And you could determine that in your calculator as well. You could plug in for the X value negative one, plug in inverse tangent of negative one and you'll get negative pi divided by four. Make sure you're in radiance and you might get something that's like a negative 0.7853 and so on. But you can also just type into your calculator negative pi divided by four and you should get that same negative 0.7853 and so on. So those are equivalent. And then another point we can identify is the X value of positive one. We're going to have a positive Y value of pi divided by four. Now, for our transformations that are taking place, we are going to be multiplying our Y values by A A is what's being multiplied in front of our inverse tangent of X. So here A equals negative five. Now we're going to take our Y values and multiply those by A. So if we had one point of one pi divided by four, we take this A or this Y value of pi divided by four. And we're going to multiply that by A which is negative five. And then we would have still a one for X value. But now we have a positive five pi excuse me, a negative five pi pi multiplied by negative five is negative five pi divided by four. So there is one of our new points. Now let's look at the other point where we know the origin is not going to change when we multiply zero by negative five. But with this negative one, negative pi divided by four, we're multiplying the Y value negative pi divided by four by negative five, the X value remains negative one and the Y value is now a positive five pi divided by four. Those are two points we can use to graph. Now we can also take a look at our asom totes. We have horizontal asymptotes at Y equals pi divided by two and Y equals negative pi divided by two. We're also going to multiply those by our A value by negative five. And so we'll now have horizontal asymptotes at Y equals pi divided by two, multiplied by negative five is negative five pi divided by two. And likewise, we have Y equals negative pi divided by two, multiplied by negative five is a positive five pi divided by two. So those pieces of information can give us specificity for our graph keeping in mind that we're reflecting it over the Y axis or the X axis, excuse me. And that there is a vertical stretching So I'm going to move these concrete values to a different part of the screen where we're going to have a blank graph that we can fill in with our transformations. So now we have a graph still have a vertical Y axis and a horizontal X axis. What's shown for the domain for the X axis is from negative eight to positive eight. And what's shown for the range for the Y axis is from negative three pi to positive three pi. So we know that this is going to pass through the origin at 00. We know we're also going to have a point at one negative five pi divided by four. So from the origin, we go to the right one unit and then go down five pi divided by four units which is going to be halfway between negative five pi divided by two. So that's about here. And then we've got another point at negative 15 pi divided by four. So from the origin, we'll go to the left one and then we'll go up five pi divided by four units, which is approximately here. And then we have our asom totes at Y equals negative five pi divided by two from the origin. We go down five pi divided by two units and we can draw a horizontal ASOM tote and then another ASOM tote at Y equals five pi divided by two from the origin. We go up five pi divided by two. And we can draw that horizontal ASOM tope. Now, we want to try to connect these as smoothly as possible. So we're going to head head toward negative infinity along the horizontal as er head toward negative infinity for the X values along the horizontal asom tote Y equals five pi divided by two. So that's up in the second quadrant and then it's going to curve through our point, all of our points that we've identified then curbing down into the fourth quadrant. And we're going to head toward positive infinity for the X values along our horizontal Asymptote Y equals negative five pi divided by two. And there's our graph indicating the vertical stretching and the reflection over the X axis. So the last thing we need to do is identify our domain and range in interval notation. The domain is all the possible X values. And here the domain remains from open parentheses, negative infinity to positive infinity, closed parentheses. And the range is going to be all of the possible Y values. And here that has changed, that's going to be from open parentheses, negative five pi divided by two, two, a positive five pi divided by two closed parentheses. And there is our graph and our domain and our range four F of X equals negative five inverse tangent of X. Well done. We'll catch you on the next one.

Key Concepts

Inverse Trigonometric Functions

Transformations of Functions

Domain and Range

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