How Do You Know if F if Continuous
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Section two-9 : Continuity
Over the terminal few sections we've been using the term "overnice enough" to ascertain those functions that we could evaluate limits by just evaluating the role at the signal in question. Information technology's now fourth dimension to formally define what we mean by "nice enough".
Definition
A function \(f\left( ten \correct)\) is said to be continuous at \(10 = a\) if
\[\mathop {\lim }\limits_{ten \to a} f\left( ten \right) = f\left( a \right)\]
A role is said to exist continuous on the interval \(\left[ {a,b} \right]\) if information technology is continuous at each point in the interval.
Note that this definition is also implicitly assuming that both \(f\left( a \right)\) and \(\mathop {\lim }\limits_{ten \to a} f\left( x \right)\) exist. If either of these do not exist the office will not be continuous at \(x = a\).
This definition tin can be turned around into the following fact.
Fact I
If \(f\left( ten \right)\) is continuous at \(ten = a\) and so,
\[\mathop {\lim }\limits_{x \to a} f\left( ten \correct) = f\left( a \right)\hspace{0.5in}\mathop {\lim }\limits_{10 \to {a^ - }} f\left( 10 \right) = f\left( a \right)\hspace{0.5in}\mathop {\lim }\limits_{x \to {a^ + }} f\left( ten \right) = f\left( a \right)\]
This is exactly the aforementioned fact that nosotros starting time put down dorsum when nosotros started looking at limits with the exception that nosotros have replaced the phrase "nice plenty" with continuous.
It's nice to finally know what nosotros mean by "nice enough", notwithstanding, the definition doesn't really tell usa just what it means for a role to be continuous. Let's take a await at an example to assist united states of america empathize merely what it means for a function to be continuous.
Example 1 Given the graph of \(f\left( x \right)\), shown beneath, determine if \(f\left( ten \right)\) is continuous at \(x = - ii\), \(10 = 0\), and \(x = 3\).
Prove Solution
To reply the question for each point we'll need to get both the limit at that betoken and the office value at that point. If they are equal the function is continuous at that betoken and if they aren't equal the role isn't continuous at that point.
First \(10 = - 2\).
\[f\left( { - ii} \right) = 2\hspace{0.5in}\mathop {\lim }\limits_{x \to - ii} f\left( x \right)\,\,\,{\mbox{doesn't exist}}\]
The part value and the limit aren't the same and then the function is not continuous at this point. This kind of discontinuity in a graph is called a bound discontinuity. Spring discontinuities occur where the graph has a break in information technology every bit this graph does and the values of the office to either side of the break are finite (i.e. the office doesn't go to infinity).
Now \(10 = 0\).
\[f\left( 0 \correct) = 1\hspace{0.5in}\mathop {\lim }\limits_{10 \to 0} f\left( 10 \right) = ane\]
The function is continuous at this point since the function and limit have the same value.
Finally \(x = three\).
\[f\left( iii \right) = - i\hspace{0.5in}\mathop {\lim }\limits_{ten \to 3} f\left( 10 \correct) = 0\]
The function is not continuous at this signal. This kind of discontinuity is called a removable discontinuity. Removable discontinuities are those where there is a hole in the graph as in that location is in this example.
From this case nosotros tin can go a quick "working" definition of continuity. A function is continuous on an interval if nosotros tin draw the graph from get-go to finish without ever once picking up our pencil. The graph in the last example has only 2 discontinuities since there are but two places where nosotros would have to pick upwardly our pencil in sketching it.
In other words, a part is continuous if its graph has no holes or breaks in it.
For many functions it'southward easy to determine where information technology won't be continuous. Functions won't be continuous where we have things like segmentation past zero or logarithms of zero. Let's accept a quick look at an instance of determining where a function is non continuous.
Instance 2 Determine where the function below is not continuous. \[h\left( t \right) = \frac{{4t + 10}}{{{t^2} - 2t - 15}}\]
Prove Solution
Rational functions are continuous everywhere except where we take division by zip. And then all that we need to is make up one's mind where the denominator is zero. That's like shooting fish in a barrel enough to determine by setting the denominator equal to aught and solving.
\[{t^ii} - 2t - xv = \left( {t - five} \right)\left( {t + 3} \right) = 0\]
And then, the function will non be continuous at \(t=-iii\) and \(t=5\).
A nice result of continuity is the post-obit fact.
Fact ii
If \(f\left( x \right)\) is continuous at \(10 = b\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = b\) and then,
\[\mathop {\lim }\limits_{x \to a} f\left( {g\left( 10 \correct)} \right) = f\left( {\mathop {\lim }\limits_{ten \to a} g\left( x \right)} \right)\]
To see a proof of this fact meet the Proof of Diverse Limit Properties section in the Extras chapter. With this fact we tin now practice limits like the following example.
Example 3 Evaluate the following limit. \[\mathop {\lim }\limits_{x \to 0} {{\bf{e}}^{\sin 10}}\]
Show Solution
Since we know that exponentials are continuous everywhere nosotros can use the fact above.
\[\mathop {\lim }\limits_{10 \to 0} {{\bf{e}}^{\sin x}} = {{\bf{e}}^{\mathop {\lim }\limits_{x \to 0} \sin 10}} = {{\bf{east}}^0} = ane\]
Some other very squeamish consequence of continuity is the Intermediate Value Theorem.
Intermediate Value Theorem
Suppose that \(f\left( x \right)\) is continuous on \(\left[ {a,b} \right]\) and allow \(K\) be any number betwixt \(f\left( a \right)\) and \(f\left( b \right)\). Then in that location exists a number \(c\) such that,
- \(a < c < b\)
- \(f\left( c \correct) = M\)
All the Intermediate Value Theorem is really maxim is that a continuous function will take on all values betwixt \(f\left( a \correct)\) and \(f\left( b \right)\). Below is a graph of a continuous function that illustrates the Intermediate Value Theorem.
As nosotros can run across from this image if we pick whatever value, \(Chiliad\), that is between the value of \(f\left( a \right)\) and the value of \(f\left( b \right)\) and draw a line direct out from this signal the line will hit the graph in at least one point. In other words, somewhere between \(a\) and \(b\) the function volition take on the value of \(M\). Also, as the figure shows the function may take on the value at more than than one place.
Information technology'south also important to annotation that the Intermediate Value Theorem just says that the part will take on the value of \(M\) somewhere between \(a\) and \(b\). It doesn't say just what that value will be. It simply says that information technology exists.
And then, the Intermediate Value Theorem tells us that a office will accept the value of \(M\) somewhere between \(a\) and \(b\) but it doesn't tell usa where it volition take the value nor does information technology tell us how many times it volition take the value. These are important ideas to recall near the Intermediate Value Theorem.
A prissy use of the Intermediate Value Theorem is to prove the existence of roots of equations as the following instance shows.
Example iv Prove that \(p\left( x \right) = two{x^3} - five{x^ii} - 10x + five\) has a root somewhere in the interval \(\left[ { - 1,2} \correct]\).
Evidence Solution
What we're really asking here is whether or not the function will have on the value
\[p\left( x \right) = 0\]
somewhere between -1 and ii. In other words, we want to prove that there is a number \(c\) such that \( - 1 < c < 2\) and \(p\left( c \right) = 0\). Nevertheless if we define \(M = 0\) and admit that \(a = - 1\) and \(b = 2\) we can run into that these 2 condition on \(c\) are exactly the conclusions of the Intermediate Value Theorem.
And then, this problem is set upward to use the Intermediate Value Theorem and in fact, all nosotros demand to do is to testify that the part is continuous and that \(K = 0\) is betwixt \(p\left( { - 1} \right)\) and \(p\left( 2 \right)\) (i.e. \(p\left( { - 1} \right) < 0 < p\left( ii \correct)\) or \(p\left( 2 \right) < 0 < p\left( { - 1} \right)\) and we'll be done.
To practice this all we need to do is compute,
\[p\left( { - ane} \correct) = 8\hspace{0.5in}\hspace{0.25in}p\left( ii \correct) = - 19\]
And then, we have,
\[ - 19 = p\left( ii \correct) < 0 < p\left( { - ane} \right) = 8\]
Therefore \(Chiliad = 0\) is between \(p\left( { - one} \right)\) and \(p\left( 2 \right)\) and since \(p\left( x \correct)\) is a polynomial it's continuous everywhere and then in particular it'south continuous on the interval \([-1,2]\). And then by the Intermediate Value Theorem at that place must exist a number \( - 1 < c < ii\) so that,
\[p\left( c \right) = 0\]
Therefore, the polynomial does take a root betwixt -1 and 2.
For the sake of completeness here is a graph showing the root that nosotros just proved existed. Note that we used a figurer plan to actually observe the root and that the Intermediate Value Theorem did not tell usa what this value was.
Let's take a await at some other example of the Intermediate Value Theorem.
Instance 5 If possible, determine if \(f\left( x \correct) = 20\sin \left( {x + iii} \right)\cos \left( {\frac{{{x^two}}}{ii}} \correct)\) takes the following values in the interval \([0,v]\).
- Does \(f\left( x \right) = 10\)?
- Does \(f\left( 10 \right) = - ten\)?
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Testify Word
Okay, and so as with the previous example, we're being asked to determine, if possible, if the part takes on either of the 2 values above in the interval [0,5]. First, let'southward notice that this is a continuous function and so nosotros know that we tin can use the Intermediate Value Theorem to do this problem.
Now, for each role we will let \(M\) be the given value for that part and then we'll demand to show that \(M\) lives between \(f\left( 0 \right)\) and \(f\left( five \right)\). If it does, and so nosotros can use the Intermediate Value Theorem to prove that the function will take the given value.
So, since we'll need the two office evaluations for each part let's give them here,
\[f\left( 0 \right) = 2.8224\hspace{0.5in}\hspace{0.25in}f\left( 5 \right) = 19.7436\]
Now, allow's take a expect at each office.
a Does \(f\left( x \right) = x\)? Testify Solution
Okay, in this case we'll define \(K = x\) and nosotros tin can run into that,
\[f\left( 0 \right) = two.8224 < 10 < 19.7436 = f\left( 5 \right)\]
Then, by the Intermediate Value Theorem there must exist a number \(0 \le c \le 5\) such that
\[f\left( c \right) = x\]
b Does \(f\left( x \right) = - 10\)? Evidence Solution
In this function we'll define \(M = - x\). We now have a problem. In this part \(One thousand\) does not alive between \(f\left( 0 \right)\) and \(f\left( 5 \right)\). So, what does this mean for us? Does this mean that \(f\left( x \right) \ne - 10\) in \([0,5]\)?
Unfortunately for us, this doesn't mean anything. It is possible that \(f\left( x \correct) \ne - 10\) in \([0,v]\), but is it also possible that \(f\left( x \right) = - ten\) in \([0,5]\). The Intermediate Value Theorem will only tell usa that \(c\)'s will exist. The theorem will NOT tell us that \(c\)'s don't exist.
In this case information technology is not possible to determine if \(f\left( x \right) = - 10\) in \([0,five]\) using the Intermediate Value Theorem.
Okay, every bit the previous example has shown, the Intermediate Value Theorem will non e'er exist able to tell united states of america what we want to know. Sometimes nosotros tin can use it to verify that a function volition have some value in a given interval and in other cases nosotros won't be able to use it.
For abyss sake hither is the graph of \(f\left( ten \right) = 20\sin \left( {x + 3} \correct)\cos \left( {\frac{{{x^2}}}{two}} \right)\) in the interval [0,v].
From this graph we can see that not just does \(f\left( 10 \right) = - x\) in [0,5] it does so a full of four times! Also notation that as we verified in the first part of the previous example \(f\left( 10 \right) = ten\) in [0,5] and in fact it does and so a total of three times.
And so, remember that the Intermediate Value Theorem will only verify that a function will have on a given value. It will never exclude a value from beingness taken by the office. Also, if we can employ the Intermediate Value Theorem to verify that a function will take on a value information technology never tells us how many times the function will accept on the value, information technology only tells us that it does take the value.
Source: https://tutorial.math.lamar.edu/classes/calci/continuity.aspx
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