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We are going to see in this video how to study the continuity of a function

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and above all we are going to see examples of functions defined in pieces.

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When is a function continuous in a point?

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A function is continuous in a point if these three conditions are met.

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First, there has to be the limit of the function when x tends to that point.

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Remember that in order for the limit of the function to exist in a point,

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There must be the two lateral limits, the lateral limit on the right,

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the lateral limit on the left, and coincide, and the result is a finite number.

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We are going to call that limit L. On the other hand, the function must be defined at that point.

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That is, that point A has to belong to the domain of the function.

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In other words, there is F in A, there is the image of A.

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And, on the other hand, the two values, the limit and the value of the function in the point, have to coincide.

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That is, in this case, as we have set the limit L,

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what the function in A is worth has to coincide with that value that we have obtained for the limit.

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For a function to be continuous in an interval, it has to be continuous in all the points of the interval.

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If it is continuous in all the points of the interval, it will be continuous in the interval.

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And many times the expression is used, a function is continuous in all its domain.

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What does that mean?

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Well, that these conditions are fulfilled for all the points that make up their domain.

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Look, this function f is a function defined in pieces.

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Notice that for x less than or equal to 3, it is defined as a rational function.

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This is a hyperbola in this case.

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and for those greater than 3, it is also defined as another interval, as another rational function.

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Here we have the graph of the function to understand what we are going to see,

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but in reality they can ask me to study the continuity of the function and not give me the graph.

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How would we do it? If they ask me to study the continuity of the function,

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Well, the first thing we are going to do is to study each of the open intervals.

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First from minus infinity to 3, see what happens, and from 3 to infinity.

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Look at the function.

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In the interval that goes from minus infinity to 3, the function is defined in this way, right?

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And this function, we see that when it is a rational function, it is not defined for x equal to 2.

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We also know that it is a hyperbola, that for x equal to 2 we have a vertical asymptote.

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What we see is that x equal to 2 is in this interval.

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Therefore, we know that here, in 2, the function will have that asymptote.

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So, this function, 2 divided by x-2, is continuous in the interval from minus infinity to 2 union from 2 to 3.

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In x equal to 2, the function is not defined, that is, there is no f in 2, the image of 2,

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and the lateral limits on the right and on the left of the function are divergent.

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In this case, the function, look, if I take a value close to 2 on the right, 2.000-2,

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the one above is going to be positive, the one below too, this tends to be more infinite.

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If you look at the graph, I can see perfectly that when I get closer to 2 on the right,

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the function goes to plus infinity, in this asymptote that it has, in x equal to 2.

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But taking limits I can also see it.

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And the limit when x tends to 2 to the left,

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in this case we would take a value close to 2 to the left that can be 1.999.

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If we make that difference in the denominator,

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will give us negative, positive between negative, negative, this limit tends to

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minus infinity. And we see it in the graph here, that when I get closer to 2 on the left

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I go to minus infinity. Well, let's analyze now the interval

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that goes from 3 to infinity. The function, we said,

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is also defined as a rational function and also

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Notice how the degree of the numerator is 1 and the degree of the denominator is also a hyperbola.

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Well, the function 3x divided by 2x minus 3 is not defined for what value of x?

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Well, when x is 3 halves, which is when the denominator is nullified.

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But what happens is that 3,5 is not in this interval from 3 to infinity,

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but it is in the interval that goes from minus 3, that is, from minus infinity to 3.

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This function is continuous in the interval from 3 to infinity.

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In this interval it is continuous.

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And we see it in the graph, in the graph, this piece of hyperbola is

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for the x greater than 3, and we see that from 3 the function is contained.

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It can be drawn from a single line.

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And now let's see what happens in x equal to 3.

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In x equal to 3, first we see what the function in 3 is worth,

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we would have to look here, which is where it is defined,

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and substituting for x equal to 3 I obtain the value 2.

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The limit, when x tends to 3 on the right of the function,

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of course, if I get closer to the right of the function, I have to take this expression for f

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and substitute in it the x for 3,

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then I would have 9 divided by 6 minus 3,

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that is, this result would be 3.

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And the limit, when x tends to 3 on the left,

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In this case, if I get closer to 3 on the left, what expression do I have to take?

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Well, I get closer to this other side, I have to take this one, for the f.

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And substituting, the result is 2.

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This, that we do here with the calculation, well, with the graph we saw it clearly, right?

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If I get closer to 3 on the right, the images of these x are approaching the value of 3.

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That's why the lateral limit on the right has given me 3.

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On the left, we also saw it very well,

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the images of these x are approaching the value of 2.

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That's why this limit is worth 2.

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And what is worth the function in 2 is defined.

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You see that here we have a full point, so for 3 the function is worth 2.

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But now we are assuming that we do not have the graph,

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because we are doing everything with the mathematical expression of the function.

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The most convenient thing is to put this diagram to see if I go to the right what function I have to place,

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for the f what expression, if I go to the left what other expression I have to put.

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In short, what do we see? That the function is not continuous in x equal to 3.

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In x is equal to 3, the function is not continuous and the discontinuity it presents is not continuous.

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The discontinuity it presents is finite jump because we see that the lateral limits are different.

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So it is an inevitable discontinuity of finite jump.

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We are going to see another example of continuity study and we are also going to add the derivability of the function.

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First the continuity and then we are going to study the derivability of the function.

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First we are going to see the intervals that we have.

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The intervals that we have, notice that the function has like three sections.

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The first one, from minus infinity to zero, is defined as x squared.

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From zero to three, the function is defined as x.

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And from three to infinity, it is defined as a constant function that is always valid.

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Well, the first thing we see is what happens in the open intervals from minus infinity to zero.

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In this interval, the function f is continuous,

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the function y equal to x squared is continuous in all r.

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So, in particular, it will be continuous in this open interval from minus infinity to zero.

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From zero to three, the function f is also continuous in that open interval.

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It is continuous because the function y equal to x is continuous in all r.

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It has no discontinuity problem.

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And from 3 to infinity, the function is defined as y equals 1, which is a constant function,

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so it is continuous, since this function is continuous in all of R.

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We still have to study what happens in x equals 0 and in x equals 3.

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Let's start with x equals 0.

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To see if the functions continue in x equal to 0, the first thing I have to see is what the function in 0 is worth.

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The function in 0 is defined here, that the inequality appears with the equal,

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less or equal, for x equal to 0 I will have to substitute here.

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In this case, f in 0 is worth 0.

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Now we are going to see the lateral limits on the right and on the left of 0.

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If I approach 0 on the right, the expression I have to take is

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y equals x, that is, here in the limit, instead of f , I will put x.

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And when I substitute x for 0, this limit gives me 0.

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And on the left, the limit when x tends to 0 on the left,

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now I have to take this other expression,

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When I substitute the x by 0, this limit gives me 0.

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What do I see?

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That these three things, the value of the function in the point, the lateral limits,

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are finite and coincide.

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So in x equal to 0, the function is continuous.

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f is continuous.

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Now we are going to see the point x equal to 1, x equal to 3.

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In x is equal to 3, we do the same, we study the value of the function in 3.

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In this case, notice that the function is not defined in x is equal to 3,

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because here it puts a strict inequality, it puts x less than 3,

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so here I can't substitute, but here neither.

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So it doesn't exist, the function is not defined in 3.

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Only for this reason the function would not be continuous,

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But if we want to classify the discontinuity, we will have to see the lateral limits on the right and on the left.

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We study the limit when x tends to 3 on the right of the function, and in this case we have to substitute,

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we go to 3 on the right, we have to substitute here, in the function that is constant.

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This limit gives me 1.

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And the limit, when x tends to 3 on the left,

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well, now I would have to take this other expression, the x, right?

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And by substituting x by 3, I get 3.

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Notice, the function does not exist in 3.

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And we see that the lateral limits do exist.

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They are convergent, one tends to 1, one is 1 and the other is 3.

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Here what we see is that the lateral limits do not coincide.

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So, what we have in x is equal to 3 is an inevitable discontinuity type.

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And the jump of the discontinuity, notice that we go from having, from being 1 to being 3.

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The jump of two units, right? Finite jump.

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The limits have come out finite and the difference between them in absolute value gives me a finite jump.

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So we have an inevitable discontinuity of finite jump.

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Summing up, the function is continuous, let's put it here, f is continuous in all r

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excepting the value x is equal to 3, where the function, in addition to not being defined,

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We have seen that the lateral limits do not coincide, there is an inevitable discontinuity of the finite jump.

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Well, seen first the continuity, we are going to see now the derivability of the function.

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We are going to study where the function is derivable.

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We have to derive the function and we are going to derive each of these expressions.

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The derivative of x squared is 2x, this would be defined for the x less than 0.

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The derivative of x is 1.

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Notice that here what we are going to put is going to be the open interval, okay?

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We are going to write the two inequalities without the equal.

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Here was the equal, here we are not going to put it, now we explain why.

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And the derivative of 1 is 0.

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That is, I am deriving the function into pieces, section by section,

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and if in any of the inequalities an equal appears, here we are going to remove it.

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We are going to remove it because then we are going to see what happens in x is equal to 0 and in x is equal to 3

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and the derivative will exist if the lateral derivatives in those points exist and coincide.

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Therefore, we a priori here we have to write the function with the strict inequalities, without equal.

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Well, another important thing is that the function where the function is not continuous

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will not be derivable, obviously, so that's why we say that before you have to study

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the continuity and we said before that in x equal to 3 the function f of x is not continuous

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So, we already know that the function in x is equal to 3 is not derivable, there is no derivative in 3.

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And it is logical, right? In x is equal to 3, if the function is not even defined,

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but if it is not even continuous, I cannot establish a tangent line to the curve at that point,

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then it will not be derivable.

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Let's go step by step, as we have done before in the continuity.

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In the interval that goes from minus infinity to zero, the function f is going to be derivable.

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From zero to three, the function is defined as x and it is derivable in all the points.

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Remember, this is the function f , in all these points I can draw the tangent line of the curve in those points.

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And in fact here, since it is a line, the tangent line would coincide with the same line.

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So f is derivable in that interval.

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And from 3 to infinity, the function was defined with a constant, right? It's always worth 1.

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I can always draw the tangent line in each of these points, so the function is derivable.

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So it is derivable in the open intervals from minus infinity to zero, from zero to three, and from three to infinity.

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What happens in zero and what happens in three?

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Well, in three we have already said that since the function is not continuous, it is not derivable.

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We have already committed this.

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Let's see what happens in 0.

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In x equal to 0, we study the derivative on the right of 0 and on the left.

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If these two lateral derivatives exist and coincide, then the function will be derivable in 0.

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In 0, if I approach from the right, note that this derivative is 1.

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And if I get closer to the left, I would have to substitute in this other expression, 2 by 0, 0.

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What we see is that the lateral derivative on the right and on the left does not coincide.

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Therefore, there is no derivative in 0.

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We have commented this on other occasions, and it is that if we have angular points, as occurs in this case,

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the function is not going to be derivable.

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So, let's recapitulate.

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The function nx is equal to 3 was not continuous,

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so it is not derivable.

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That's why we have to study continuity before derivability.

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And nx is equal to 0, there is no derivative in 0,

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because if we study the lateral derivatives on the right

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and 0 on the left give us different results.

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That is, if I draw the slope of the tangent line to the x-squared curve,

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I would get a slope of 0,

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but if I draw it on this other side, on the left,

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I would get slope 1.

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That is, several slopes cannot exist.

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If there is a derivative, it is unique.

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So there is no derivative of 0.

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f of x, therefore, is continuous in all r, minus in 3.

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And it is derivable in all r, minus in 0 and in 3.

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In 3 because the function is not continuous, in 0 because the lateral derivatives do not coincide,

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and then the derivative does not exist.

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Let's see one last example in which they give me a function in pieces,

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in which some parameters a and b appear and they ask me how much a and b have to be worth

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for the function to be continuous. We are going to analyze, as always, first what happens in the

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open intervals. In this case we see that the function from minus infinity to minus 2 is defined

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as a straight line, a polynomial function of grade 1, from minus 2 to 3 it is defined

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with a constant function equal to 4, and from 3 to infinity the function is defined as another

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slope-dependent line, ax-2. We first study the open intervals, from minus infinity to minus 2.

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In this first interval, the function is continuous, and it is continuous because the function equal to 3x plus b is continuous in all r.

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In the open interval from minus 2 to 3, the function also does not present any continuity problem,

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is continuous also because y equal to 4 is continuous in all r.

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And from 3 to infinity, f of x is continuous also

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because y equal to x minus 2 is continuous in all r,

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therefore it is going to be continuous in this open interval.

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I want the function to be also optimal.

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For this we are going to see what the function in minus 2 is worth.

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And in minus 2 we have to substitute here.

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Here it is defined for minus 2, because the image of minus 2 is worth 4.

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The limit, when x tends to minus 2 on the right,

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then I have to substitute, if I get closer to minus 2 on the right,

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here in this function, which is going to be a constant, I have 4 left.

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And the limit when x tends to minus 2 on the left,

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in this case if I get closer to the left of minus 2, I have to take this other expression,

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3x plus b. When substituting for minus 2, this would be minus 6 plus b.

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I want the function to be continuous in x equal to minus 2,

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therefore the three things have to exist and coincide.

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In this case, 4 has to be equal to minus 6 plus b.

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So from here we clear that b has to be worth 10.

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Now let's see what happens in x equal to 3.

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In x equal to 3, the function is defined for x equal to 3, it is worth 4.

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So f of 3 is equal to 4 as well.

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And we study the lateral limits by the right side of 3 and by the left side of 3.

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If I approach the right side of 3, the expression I have to take for f would be ax-2.

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And the result of substituting would be 3a-2.

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The limit when x tends to 3 on the left, on the left of 3, the function f of x is defined as the constant function 4.

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So this limit is 4.

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And I want the function to be continuous 3.

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So it has to be 3a-2, it has to be 4.

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Resolving, I have to have a, it has to be 2.

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In this case, I got a system of two uncoupled equations,

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in the sense that from one of the equations I have cleared the b

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and from another of the equations I have cleared the other parameter, the value of a.

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Sometimes I get a system of equations that I have to solve.

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Well, if a is 2 and b is 10, those are the values ​​that a and b have to take

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for the function to be continuous throughout the row.