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The function satisfies f(r + 2) f(r) for all €. Use substitution to show thatf(r) dc = 2 f(z) dc....

Question

The function satisfies f(r + 2) f(r) for all €. Use substitution to show thatf(r) dc = 2 f(z) dc.

The function satisfies f(r + 2) f(r) for all €. Use substitution to show that f(r) dc = 2 f(z) dc.



Answers

Show that if $f$ is any function, then the function $E$ defined by $$E(x)=\frac{f(x)+f(-x)}{2}$$ is even.

Okay, So the first thing if something's uneven function that means f of X is equal to 1/2 of negative X and G of X is equal to negative G of negative axe. That's just the definition of our even functions. S O f of two is to means f of negative two is also to. And if g up to his negative to that means g of negative too Negative too, is the f of g of to We're told that we're told that GF two is negative two. So it's gonna be half a negative too, which is just equal to two. And g uh, have a negative, too is equal to G 02 which is equal to negative two. Okay.

Come on. And so from your question, you know why is he going to f. Of you and you? It's equal to G. Of X. So from here we know. Dy open the X. Can be written us and do Y over to you. Nice. Do you over the X. So this is by using the chain rule. Yeah. Mhm. So as we can see when the U. Device and you are going to be left with do I over the X. So we've not actually done anything to it. And so from here. So using this um chin rheumatic, we are going to differentiate again with respect to X. So we have this of do I over the X. So this is going to be equal to differentiating what we had earlier with respect to X. So we have the Y. Already you nice steel over the X. So from here we can see we're going to get described why? Over the X. Squared. So no, we have this again. Yeah. Mm hmm. Sure. So now differentiating, we are going to get we have the Y. Okay, do you thanks the U. So D. Do you over deep the X. Then different shitting again with respect to X. Over here then. Plus we have to you over the X. Thanks. Um D. Over the Y. Over the U. Then we differentiate with respect to our X. Over here. So from here we are going to get Dy where do you okay, described you over the eggs word. I'm A. Plus do you over the X. Times. Um Mhm. So we have uh D. Mhm. So you'll learn to differentiate this just I'm with you worry over to you over here. With respect to the you then we're going to multiply this by the U. Over tax. So from here we are going to get the Y. So we have dy okay over the X. Sorry over do you thanks disk. Where'd you? Yeah. Over the X. Word mm. Plus. So we have class over here then he has to you over the X. Yeah. All multiplied by. We have described Why over to you squared. Thanks do you over the X. So from here you're going to get we have the Y. Was that you times described you over the X squared? Bless. Do you go by the X christ? E. You over the X. Oh just miles have played by. Disquiet the way what do you square? So from here we are going to get to during this side you're going to get described why? Over the U. Squared? All my multiplying by. Do you over the X. I'll sweat plus. So we are going to get dy over to you. Thanks. D. Squared you by the X. Square. So therefore yeah. So therefore um described why over the X. Word is equal to described why over the use word multiplied by. Do you look at the expo sweat plus. Do you worry already? You thanks disk. Where'd you? Over the X. Squared? Yeah

Okay, so we are asked to show that the integral off E equals R C divided by two. So this pretty easy. Just have a, like, a small trick. But we know that the integral off e to the constant, uh, anti. Okay, is the one divided by the constant on just the same value I have off e plus C. Okay. And, uh, as I'm going to do this, you have to remember, this is for indefinite. Integral. But I actually have some values for the limit. So the integral off zero infinite off E to the minus two. T divided by r c. Did she must be equal. Okay, so one over the integral that is minus two times two divided by R. C. Okay, I want to play by E on exactly the same I had before, minus two. See, divided by R C. Okay, but this is going to be value from zero. So uh huh. Infinity. Okay. So when I do this, I can just, uh you know, this, uh, turns around because he's divided. One is divided by this. So this is minus RC, divided by two. Okay. And I have e. Okay, so the two minus two RCTI multiplied by infinity. Okay, which is just on infinite on then I have Ah, this minus e 20 Because when I multiply zero by the constant, I just get zero. Okay, so this is the little bit off a trick, actually, is pretty easy. I know that anything divided, I mean anything to the zero. It's actually one. Okay, so this value is going to be one onda uh, what happens when I have, uh, something, uh, elevated to the infinity. And actually, this is because I have ah, minus sign here. I have to the minus infinity. So if actually we just see at the blood of E, we know that, uh, finitely positive is to the right on minus infinity. It's to the left. And I see that Asai approach minus infinity. Mhm. So the X tends to zero. So actually, these values just going to be zero. Okay, so these councils and I just have this value that is one. So I have ah minus r c divided by two. I want to play by minus one. This just changes the same. So I just have RC divided by two. So with this, I just made the thing. I approved the thing I wanted to do

Hello, everybody. In this video, I'm gonna be showing you how to solve exercise 19 in Chapter one, Section one of calculus, Early transit Dentals. Now, this problem tells us that we have two functions f N g. And they tell us that f is even and the f of two equals two. They also tell us that G Izod and G of two equals negative too. And then they ask us to evaluate the four different expressions below. They're the most important. Two things to acknowledge before we start is that because f is even, we have that for any X f of negative X equals f of X. And because G is odd, we have that for any X g of negative X equals negative g of f and this is all the information we need. Let's start with f of negative, too, using the expression that we have here. From the evenness of it, we see that negative too. It's a sequel to effort to and from this expression here we know exactly what that is. That is just to similarly Fergie of negative, too. We have from this expression here that you have negative too. It's just negative g of to And since G of two is negative two, we have that This expression equals positive too. The last two questions have to do with compositions, which makes it slightly more complicated. But the same procedure can be used and then it is still not very hard at all for f of g of to We know that g of to from this expression appear is just negative too. So this turns into f of negative too Have been using the same logic is in the first problem. We knew that f of negative too is just f of two from its evenness just to similarly, for the last problem. We knew that f of negative too is just effort to which is to And then we know that g of to negative too. So that is how you solve exercise 19


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