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) = sin⁻¹ x + π/2

Accepted Answer

Welcome back everyone. In this problem, we want to transform the graph of Y equals the inverse of sine X to graph the function F of X equals the inverse of sine X plus three halves of pi. Our original domain is between negative 1 to inclusive and the range is between negative half of pi to half of pi inclusive. When we do our new graph, we'll want to rewrite the domain and the range with interval notation. So what do we already know? Well, for starters this function F FX is one function that has been expressed as another or, or shifted by another recall that if we have a function that's equal to some other function, let's call it G of X plus a constant is called K, then that constant represents the vertical shift. So in this case, all vertical shift is three halves of pi. So what this function is basically telling us is that we're going to graph the inverse of sine X and then we're going to shift it upwards by three halves of pi. The good thing about that is that we already have the inverse of sin drawn for us So all we need to do now is recreate the graph and then shift it up. So let's do that down here. So first let's draw back the inverse of the sign of X. So remember it was between negative one, which would be right here. Let me like that in a darker color and positive one. OK. And it passed through the origin. So the graph is gonna look something like this. OK. It's my, this is my best attempt, don't judge me. So we have our graph already. This is the graph of the inverse of the sin of X. And now we're just going to shift it up by three halves of pi. So let me take the graph here and I'm gonna shift it up one unit, two units, three units because that was an upward shift of three halves of pie. OK? No, we have it at our new point. Make sure I put that here and we just need to identify what those points are here. Remember this was negative one and we had one right here. It's a vertical shift. So we didn't move it horizontally. So this point would now be negative one pi and our other point right here would be 12 pie. So we've shifted the graph, this would not represent Y equals the inverse of sine X plus three halves of pie. So finally, let's write our domain and range. So it was a vertical shift. OK. So because it was a vertical shift, our domain is still between negative one and one inclusive. However, our range has changed because no, the lowest point is pi right? And the highest value is two pi. So our range would be from pi to two pi inclusive. And that would be our graph of F FX. Let me write that properly, not Y but rather it should be F FX. Thanks for watching. I hope this helped.

Key Concepts

Inverse Trigonometric Functions

Transformations of Functions

Domain and Range

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