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.f(x) = cos⁻¹ (x + 1)

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 cosine of X to graph. The following function then identify the domain and range express the domain and range in interval notation. Our function that we are asked to graph is F of X equals the inverse cosine A X plus eight. And then we are given a graph of Y equals the inverse cosine of X. The domain for what's for our graph is from negative one to positive one inclusive of both. The range for our graph is from zero to pi inclusive of both. And then we can identify some points here. Our graph is drawn on a vertical Y axis. A horizontal X axis. It intercepts with the X axis at one zero, it intercepts with the Y axis at zero pi divided by two. And then the point which is furthest to the left and furthest up on our graph is located at negative one pie. So there are three key points for our graph. Now what transformations are taking place, we look at our function F of X equals the inverse cosine of X plus eight. The big difference here is this plus eight that's being added to our X values. Now, how does that transform it? Well, when we are adding something to our X values, that indicates a horizontal shift because it's being directly added to our X values. And because we're adding, because that is a positive number, it's going to be a horizontal shift to the left. And so specifically, we are going to be subtracting that eight. So we're subtracting eight from every X value. So we've identified our points negative one. So one at a different color, our points negative one pi zero pi divided by two and 10. But now we're going to subtract eight from each of those X values. And we'll now have for the first one, negative one minus eight is negative nine. Still pi for the Y value for the next minus eight is negative eight. Still pi divided by two for the Y value. And for the last 11 minus eight is negative seven still zero for the Y value. And now we have those three key points and we can use them to draw our graph. We'll shift this down to a different part of the screen where we'll have a blank graph. And now we still have a vertical Y axis, a horizontal X axis. They come together at the origin in the middle, the domain for what's shown of our blank graph is from negative 10 to positive 10. And the range for what's shown is from negative three pi divided by two to positive three pi divided by two. So now let's plot our three points. The first one is at negative nine pi from the origin. We go to the left nine and we go up pi the second point we have is native eight pi divided by two from the origin. We'll go to the left eight and we'll go up pi divided by two. And then the last one is at negative 70 from the origin will go to the left seven. And we want to connect these with a smooth curve. And that is our graph of F of X equals the inverse cosine of X plus eight. And the last thing we are asked to do is identify the domain and range of our graph. The domain is all of the possible X values and these are inclusive. So we put a bracket to say, say that it includes and then the first, the farthest left it can go is at negative nine and the furthest right, it can go is negative seven. So our new domain is from negative nine to negative seven inclusive of both the range. The lowest we can go for our Y values. This is also inclusive. So we have brackets that's from pie. And there is our graph and our domain and our range. 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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