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13 Suppose f is continuous function on [a, 6]. If for every continous function g on [a, 6] with g(a) 9(6) 0 it holds that Ja f()g(w)dx 0. show that f () 0 for all x...

Question

13 Suppose f is continuous function on [a, 6]. If for every continous function g on [a, 6] with g(a) 9(6) 0 it holds that Ja f()g(w)dx 0. show that f () 0 for all x € [a,6].

13 Suppose f is continuous function on [a, 6]. If for every continous function g on [a, 6] with g(a) 9(6) 0 it holds that Ja f()g(w)dx 0. show that f () 0 for all x € [a,6].



Answers

Use the result of Exercise 82 to prove that if $g(x)$ is continuous, then $f(x)=\lg (x) |$ is also continuous.

The question is Afghanis to use terms two and three and the laws of continuity to show that you the X times gives and three X is Canadians. So there to purchase submarines. There are two parts to this each of the ex, which is continuous by the room two on the real line. And the second part is co sign of three X. And here we have to use their own vibe, which talks about the continued e continuity of composite functions. Ah, three acts as a polynomial is continuous. Hi. The're, um too. And sorry. This e t actually be continued by the three. Um and co signed X continuous. Bye, dear. Um three and so Coach shine. Very ex continuous. Bye. Zero. Um, I should be saying continuous on the line or continuous everywhere for each of these. Ah, but typically, in practice, when you say continuous, you mean continuous everywhere. And if you mean continuous on a restricted range, you would specify that. Okay, now that we have each the ex being continuous in the re line and cause I had three ex being continuous on the real line, uh, we can use the basic laws of continuity, which celebrity as be Elsie, that the product of two continuous from Jones is continues and that's it.

For the following problem, let's go back and understand what it means for F and G to be continuous. We know F is continuous. If the limit as X approaches C of f of X equals F F C and the same exact thing Fergie, except we just replace the EFS with Gs. So if we want to divide these two, if we want to look at F over G. So like this, we know that algebraic lee. This is the same exact thing, right? As FFC divided by GFC. Again, this is all assuming that GFC is not ever equal to zero. So that's an assumption we have to make. But what we know about our limit rules is that if we have a limit on the top and the bottom, we can remove these, we can just apply the limit on the outside. So by doing this, we see that the limit of this whole fraction is equal to this whole fraction, which means that it's going to be continuous.

And question number 16 recalled the definition of continuity on an interval. Objection G is continuous from our table if it's continuous at every number in the interval. He called the definition of continuity at a number. Our function G is continuous at a number eight if limited G x equal to you. Okay, note that we need to show a function Z is continuous at a number eight. Mm hmm. To show that values of limit your fixing X goes to a and g f a r equal. So we must be sure that one never did you fix the justice? Two. He is defined Ah, hey, Therefore, to show that geo fix equal to X minus one over three X plastics is continuous on negative infinity So negative two we need to show Secondly, you fix equal to your faith at every hey belong to negative activity and negative two. So he must be sure. Yeah. Do you fix equal to X minus one over three X was six is defined, right? Every A belong to international community and negative two as we know, the domain of rational function consists of all the real numbers except those for the which denominator is zero. So three X plus six equal to zero three x equal. Conectiv six. So it's equal to negative two. Therefore, the domain folks Magique Yes, negative community. And to needed to in post different things. So you fix. Take it to experience 1/3 x six is the point. Ah, every I belong to negative equity and negative two. His value is your faith equipped. A minus 1/3 8 was it remains to find, and then it's geo fixed. When it goes to a it existed by using properties of limits. We call these ones which will be used some roads, different law quotient flow, which say that the limit of precaution is a caution for the limit, provided that the limit of the lemon nature is not zero for constant multiple. No, the limit of a constant times a function is the constant times the limit of the function. So we have the clinic Do you fix? It's goes to pay equal to limit. It's supposed to a four X minus 1/3 x posted equal to named excellent as well supposed to pay. So we remember three x simplistic supposed to aim by cushion flow that it minutes ex plastics not big 4 to 0 So equal to there are pics exposed to pay my Muslim one x goes to pay mhm, then three years because he goes to pay minus Let's lot six. It goes to a that by different slow and some of those. Then we got that Olympics months late, one over street and take plus there. Six. When you go to a minus one over 38 plus six plus six. Therefore, we thought that and do you fix? It's close to eight. Equal to a minus 138% Go to G with a So we got the results. That's let me get your fix. Mhm. It's close to a equal to GfK so G is continuous on negative infinity and negative two by the definition of continuity. And that's in the final answer. Thank you

So in this question were given, the problem is for his equation for f of X, which is X squared minus 36 Over X -6, Even X is not equal to six and if X is a call to sex It's going to be 13, so 13, if X is equal to six having this, we need to see if the function is continuous at a is equal to six. So in order to figure out continuing continually, we need to show that these values are equal and if of six is a call to the limit as X goes to six from the left of F a bex and also equal to the limit as X goes to six from positive over for Becks. Now, if f of we have about six then As um what is FSX is 13, we have it here. So therefore to here we have 13 for the limit as X goes to sex from the negative of f. Well, let's just rewrite it here. That is going to be the limit as X close to six from negative of X word minus 36 Or X -6, that's the definition of that. Uh active coaches checks from the left now, that is going to be the same as if I simplify it, the numerator is x minus x times X plus six And after it simplification and just get an X-plus six As X costs six. This limit becomes 12. So we already see, we don't have any equality between the first two terms. So we already know that our function is not continues at six.


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