Question
Find all critical values of the finction f(r) = 2r'-9r' 601 .Lctf()-V A student determines that for some reale numbera, f (a) - -1. Explain why this sudcnl Iasoning must bc fulty:
Find all critical values of the finction f(r) = 2r'-9r' 601 . Lctf()-V A student determines that for some reale numbera, f (a) - -1. Explain why this sudcnl Iasoning must bc fulty:


Answers
Explain how to find the critical value for an $F$ -test.
First, I have to find the critical F values in this question, and I can use either the tables or I can use some statistical Softwares like our Morita in this demo here, I'm going to use an online tool. So let me just take any of these examples. Let's say I take this one away of the second last one. 40 is the first degree of freedom. Then 30 is the second degree of freedom. And then I have 0.5 here. All right, over here, I just write this one degree of freedom as 40. What was the second one? The second one, Waas 30. The second one Waas 30 and this was 0.5 Let's try to calculate this, and the value that I get is 1.79 This is the critical value that I'm getting 1.79 So actually there should be 1.79 This is how we find the F value
Okay, so let's start by finding or folding derivative. So if you don't have X and signed numbers, so we'll be using our product roll. So derivative of X is itself. So we have been fine inverse of I, and we have plus deserve it of of sine inverse that's went over to square roots of one minus X squared times X Andi well lets you we want to know when this is equal still, So we know that when we have sine inverse of zero that gives a cell and when I exit also. So we have Joe. So we see that are critical points is at X is equipped stroll and we also have to include our endpoints and now we'll plug in our functions or our exports into our function. Okay, so we have negative one interests of negative one. We got 1.57 and then that zero we get so and at one, we have 1.57 So we see that we have absolute max at two points and then we have an absolute men when X is equal to zero
Okay, so we'll start by finding are critical points over fallen function. Let's take our derivative so derivative of X squared. That's two X and then we're taking derivative of coastline angers. That's negative. One over the square root of one minus X squared. Okay, so let's combine this into a single fraction. So that means this term I multiply it up at a square root of one minus x squared and divide by the same time. Okay, So that gives us, um, two X square roots of one minus X word minus one over the square root of one minus x squared, and I will start a sequel to their all. Okay. And now we want to solve for, um For when our extremities there also, we have our numerator. Now we'll add by one and divide by two. So we get X square roots of one minus X squared is equal to 1/2 on. And when is this true? Okay, so now it's divide by X and then take the score irritable size. So we get one minus X squared is equal to over four x squared. It's multiply X squared on both sides. So I get X squared minus except for four is equal to 1/4. And I want to do everything to our right hand side. So we have extra parts. War one is acceptable to minus or plus one of before is equal to zero. And now can with actress. Okay, so this is a sequel to, um X squared minus one house to a part of to and that you're just x squared is equal to 1/2. So x is equal to plus minus the square roots or one of this order to Okay, so now let's go back up. So we're gonna include our endpoints. Excellent. Toom minus one over square it too. X is equal to one over the square root of two, an acceptable to one. And Alex finds are falling. Um, f exploits, given our exploits. So I put this into my calculator, um, and see what I get. Okay, so I got 4.1 born and now for native one day, a lot of by the square root of two square and plus coastline. Inverse of you have one divided by the square root of two. I got 2.8 six, you know, for a positive version of that. I get 1.28 and for one, I get one self. Okay, so we see that our absolute tracks is at X is equal to negative one, and our absolute minimum without X is equal to one.
Okay, so let's start already taking are derivative to find our critical points. That's to co sign of X, an interpretive of coastline. Answer that sign of X negative sign of X and I will set this equal to their own on when is, um, X equal to zero in this case? Well, when co sign when we have close on a pirate, too. We have zero times fine of one. So we could have when X is equal to fire. But you because that makes coastline zone. You can also have when X is equal to zero on pie because that's when it makes sign is their own. Okay, so these are our critical points. So let's write them out like so have zero. However two on and hi and then let's find are falling. Outputs are ffx. Okay, so we get our fallen helps 1011 So we see that we have to Absolute Max is that's that ecstatic with zero and negative for a pie. And then we also have an absolute hman, and that's when X is equal to pi Over two