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

Question

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

b. Compute and graph f'.

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 bracket -1.1 square bracket. Awesome. So first off, in case you do not recall, the square brackets mean that this is a closed interval. So it's the closed interval of -1.1. And what we're ultimately trying to solve for in this particular problem is we're trying to create a sketch of this graph of this function of GFX on this set point within this closed interval of this per specific point of -1.1. Awesome. So, first off, in order to solve this problem, we need to determine the derivative, so we need to take the derivative of GFX is equal to X minus 1 multiplied by the inverse cosine of X. So ultimately we need to figure out the derivative of GFX, and in order to find the derivative of GFX, we need to recall and use the product. for differentiation, which, as you should recall, the product rule states that D by DX where this is a fancy way of saying we're taking the derivative with respect to X, so D by DX of U of X. Multiplied by V of X. is equal to VX multiplied by U X plus UX multiplied by V. Of X. Fantastic. And once again, that is the product rule of differentiation. So now, or rather the product rule for differentiation. So now when we solve for our first derivative for G of X, we could say that G of X denotes our first derivative for G of X. And G of X is equal to inverse cosine of X multiplied by D by D X X minus 1 plus. X minus 1 multiplied by And then we're gonna multiply this by D by DX of inverse cosine of X, fantastic. So then we can go ahead and write that G of X is equal to inverse cosine of X multiplied by 1 plus X minus 1. Multiplied by -1 divided by the square root of 1 minus X2. Fantastic. So now, our next step. we need to simplify, and when we do, we'll get that G of X is equal to inverse cosine of X minus X minus 1 divided by the square root of 1 minus X2. Fantastic. So now, our next step is we need to recall and note that the function G of X is equal to X minus 1 multiplied by the inverse cosine of X is originally defined for -1 is less than or equal to X and X is less than or equal to 1 because that because due to the domain of, so this is due to the domain of Inverse cosine of X is square -11. Fantastic. However, as we should recall a note, when differentiating G of X is equal to inverse cosine of X minus X minus 1 divided by the square root of 1 minus X2, which is we just found together. And the term X minus 1 divided by the square root of 1 minus X2 involves a square root of 1 minus X2 term in the denominator, and this square root will be 0 when X is equal to positive or -1. So let's make a note of that. So The square root of 1 minus X2 will be equal to 0. When X is equal to plus or minus 1. So making the derivative undefined at these points when it's equal to 0, and because when it's equal to 0 will be undefined at these points. So therefore, as a result, the domain of G X is strictly within this open interval of -1.1. So let's make a note of that. Parenthesis -1, 1. So now, we, in order to help us sketch our graph, we need to create points that we can plot on our graph. So let's create a quick little graph of points for us to plot on our graph to help create an accurate sketch. So when X, let's say let's find the points to graph when X is equal to -0.9, 0.2, 0.3, and 0.8. So let's find the points where when X is equal to this value, so we're going to be plugging in these different values for X into our equation. To see what our value for G of X approximate will be. So let's make a note in parenthesis that this is approximate. Which I abbreviated. Awesome. So when we plug in a value of -0.9 in for G of X, so when we plug in a value of negative 0.9 into our expression, Which I will put a green square around our derivative for G of X. So the first derivative of G of X is G X. So when we plug in a value of negative 0.9 in for X into our expression, we will get a value of 7.0495 when we round to 4 decimal places, and for negative 0.2, we will get 2.9969. And for 0.3, we will get 1.9999. And finally for 0.8, we will get 0.9768. Fantastic. So now we can use these points to plot our graph, and we need to recall and note that when plotting the points, you'll note that the G of X values can be approximated to the nearest whole number, and this will still capture the general shape of the derivative. So what we're going to do is we're going to plot these. Four different points onto our graph and then we're going to connect these 4 points with a smooth curve or a smooth line. So we're going to be drawing a smooth curve through these points to create our graph of G of X. So when we do that, we will find that our following graph, which I've gone ahead and used graphing software. To plot our graph. And as you could see, when we plot our four points, which is at -0.9.7.04946, and we have negative 0.2.2.9969 and 0.3.1.9999 and 0.8.0.97683. And when we plot those words, it looks like it's an elongated S shape almost, and all you do is we just connect our four points together with a smooth curve, and this will be our final answer for this particular problem. Hooray, we did it. We found our graph for Y equals G X, which is our final answer that we are ultimately trying to solve for. 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'b. Compute and graph f'.f(x) = (x−1) sin^−1 x on [−1,1]

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62–65. {Use of Tech} Graphing f and f'
b. Compute and graph f'.
f(x) = (x−1) sin^−1 x on [−1,1]