62–65. {Use of Tech} Graphing f and f'
a. Graph f...

Question

62–65. {Use of Tech} Graphing f and f'

a. Graph f with a graphing utility.

f(x) = (x−1) sin^−1 x on [−1,1]

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Sketch the graph of G of X is equal to X minus 1 multiplied by the inverse cosine of X on square -1.1 2. Awesome. So it appears for this particular problem, we're ultimately trying to sketch a graph of this GFX function on the point or within the domain of where this square bracket means it's closed, so a closed interval of -1.1. So with that in mind, our 1st, 1st off, we need to note that the function of G of X is equal to X minus 1 multiplied by inverse cosine of X consists of an inverse cosine function. So we need to note that it's that this cosine to the power of -1, this inverse cosine of X. Which is going to be defined at -1 is less than or equal to X and X is less than or equal to 1. Awesome, so -1 is less than or equal to X and X is less than or equal to 11. The given interval is already within the domain, so because of this fact, there will be no restrictions needed. So now our next step is we need to find points to plot on our graph because we're ultimately trying to create a sketch of this graph of G of X. In order to do that, we need to find some points to plot. And to do that, we need to evaluate GX so let's write this down. We're going to be evaluating G of X at X equals -1.0 and 1. So with that said, we need to take G of -1. Which is equal to, when we plug in a value of -1 in for X into our original function, which will be -1 minus 1. Which will be equal to and and don't forget that we need to take -1, -1, and we need to multiply it by inverse cosine. Of -1, and when we do that, we will find that G of -1 will be equal to -2 multiplied by pi. Which will give us -2 pi. which we need to note that this will be at the point. 12 pi. Fantastic. So in an effort to save time, I'm not going to write down the whole function plugging in our value of X in. I'm just going to skip that and just give you the answer. So following the same pattern, so when we're plugging in 0 for X, so G of 0 is going to be equal to negative pi divided by 2, which will mean that our point will be 0. negative pi divided by 2. Fantastic. And finally, for G of 11, it will be equal to 0, so that will mean that our point is at 1.0. Fantastic. And once again noting that for our points, it's important to note that this is in the XY format. So 1 denotes X and 0 denotes Y. Awesome. So now that we got 3 separate points, let's plot them on a graph and then draw a smooth curve between these points to graph our function. So I've gone ahead and used some graphing software to produce a digital version of our graph. And as you can see, we have our 3 separate points. So we have -1, -2 pi. And then we draw our smooth curve connecting to 0. negative pi divided by 2, and then finally at 1.0. And we just connected these three points with a smooth curve. And that's it. That is our final answer. Hooray, we did it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

62​–​65. {Use of Tech} Graphing f and f'a. Graph f with a graphing utility.f(x) = (x−1) sin^−1 x on [−1,1]

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62–65. {Use of Tech} Graphing f and f'
a. Graph f with a graphing utility.
f(x) = (x−1) sin^−1 x on [−1,1]