Using continuity to determine if there is a number one more than it's cube

[bsmithysmith]

First off, it's:

\(\displaystyle x = 1+x^3\)

Turned into function as:

\(\displaystyle f(x) = x^3 - x + 1\)

From my understanding, we need to find an interval in which x will be one more than it's cube. Giving some points, I started off with (0,1), (1,1), (-1,1), and (-2, -5).

Where I'm confused is how and where do I find the interval?

I'm also having the same difficulty with using continuity to determine if there is a solution to the equation:

\(\displaystyle x = cos(x)\)

It's just the same question: "where is the interval, how do I find it"? Because I know I can't just put random numbers and pray.

 

[Evgeny.Makarov]

Please note that the complete statement of the problem must be in the body of the post, not in the thread title.

You determined that $f(-2)=-5<0$ and $f(0)=1>0$. By the intermediate value theorem there exists an $-2<x<0$ such that $f(x)=0$, i.e., $x=x^3+1$.

Similarly, if we define $g(x)=x-\cos x$, find $a,b$ such that $g(a)<0$ and $g(b)>0$ and apply the intermediate value theorem.

 

[bsmithysmith]

Please note that the complete statement of the problem must be in the body of the post, not in the thread title.

You determined that $f(-2)=-5<0$ and $f(0)=1>0$. By the intermediate value theorem there exists an $-2<x<0$ such that $f(x)=0$, i.e., $x=x^3+1$.

Similarly, if we define $g(x)=x-\cos x$, find $a,b$ such that $g(a)<0$ and $g(b)>0$ and apply the intermediate value theorem.

I watches plenty of videos of the intermediate value theorem the previous day, and from what I understand, within the interval [a,b] with f(a)=a and f(b)=b, the value of c exist as f(c)=c.

What my teacher told me, hopefully I did hear it properly, is that a good tip is to get a negative value and a positive value close to zero. And with that equation, or question I posted, the values to determine continuity is approximately around the value zero. I may be wrong, but I'm new to the class and the summer quarter/part-time job and late night studying makes it hard.

 

[Evgeny.Makarov]

I watches plenty of videos of the intermediate value theorem the previous day, and from what I understand, within the interval [a,b] with f(a)=a and f(b)=b, the value of c exist as f(c)=c.

That's not what the intermediate value theorem (IVT) says. The value of the function at the left end $a$ of the segment does not have to be $a$ itself. It can be any $a'$, not otherwise related to $a$ except that $f(a)=a'$. Similarly, $f(b)$ can be some $b'$ not necessarily equal to $b$. You also need some value $c'$ such that $a'<c'<b'$ or $b'<c'<a'$ depending on which of $b',a'$ is greater. Then there exists a $c$ such that $a<c<b$ such that $f(c)=c'$.

This is very intuitive. If at noon you were on the highway at the 165 miles marker and at 2pm you were at the 25 miles marker, then at some point between noon and 2pm you must gave passed the 100 miles marker. Here $a=12$, $b=14$ (military time), $a'=f(a)=165$, $b'=f(b)=25$, $c'=100$ and $f$ is the function that maps time to your location. Note that $a$ does not have to be equal to $a'$; in fact, $a$ has the units of time while $a'$ has the units of distance.

What my teacher told me, hopefully I did hear it properly, is that a good tip is to get a negative value and a positive value close to zero. And with that equation, or question I posted, the values to determine continuity is approximately around the value zero. I may be wrong, but I'm new to the class and the summer quarter/part-time job and late night studying makes it hard.

Not having the full grasp of the subject is not a problem, but try to express yourself precisely. What negative and a positive values are you talking about, and they have to be close to zero for what purpose? I assume these are $a'=f(a)<0$ and $b'=f(b)>0$; then there exists an $c\in(a,b)$ such that $f(c)=0$. For the purpose of the IVT, it is irrelevant how close $a'$ and $b'$ are to zero; the important thing is that $f$ is continuous. Closeness to zero may be important for numerically finding the required $c$, but this is a different issue. Of course, if I were to find $a$ and $b$ such that $f(a)<0<f(b)$ for $f(x)=x^3-x+1$, I would not go for $b=10^5$ so that $f(b)=999\,999\,999\,900\,001$ in order not to make the reader's life harder. There is no reason not to choose $b=0$, as you did.

 

[Deveno]

Another way I have hear the IVT paraphrased is this:

Suppose there is a tree line on a mountain exactly 1000 ft. below its peak. For the sake of argument, suppose the mountain is 5,000 ft. above sea level at its peak.

If you climb the mountain from its base to its peak, at some point you are guaranteed to cross the tree line.

On a real interval $[a,b]$, the numbers $c$ between $a$ and $b$, that is: $a < c < b$ are "intermediate numbers". In the same way, the values $f(x)$ with $f(a) < f(x) < f(b)$ are "intermediate values" (of the function).

You can also think of it this way:

Imagine $f(x)$ represents the altitude, relative to sea level, at point $x$. If $a < b$ (measured say, as distance from one's base), and $f(a)$ is negative (under water), and $f(b)$ is positive (above water), then somewhere "in-between" $f(x)$ must be 0 (right at sea level).

Why is this so? Because the functions $f$ we are talking about are CONTINUOUS, which disallows "sudden jumps".

 

[bsmithysmith]

Got it:

So as Evgeny mentioned, driving 40 Miles per Hour on the highway and 25 Mile per Hour in the city, at some point in time between the highway and the city, the driver has gone 30 Miles per Hour. That's the one that stuck in my head the most.

40 is interval (b), per say, and 25 is (a)

\(\displaystyle 25 < Speed < 40\)

I've never done derivatives so I won't understand that for another week or two.

I think my teacher also made one himself. He closed the door, and asked how he was going to make it to point (b) when the door is closed. He can't, therefore it's a discontinuity.

So to sum it all up, to prove that \(\displaystyle x = cos(x)\) has a solution using continuity, we must see if it is continuous or not by turning it into a function. From there we have to find out any discontinuities (which it doesn't) so we know that x has a solution. Given a good starting point is x = 0, it would be (0,-1). Given another good input is if f(x) = 0, around (0.739,0), and that would be my intervals, -1 and 0.

So on questions such as these; would one say that x = 0 and f(x) = 0 is a good starting point to find if an equation has a solution using continuity? I may also need help with the arithmetic (a little rusted) for:

\(\displaystyle 0 = x-cos(x)\)

 

[Evgeny.Makarov]

I want to make it clear that applying the IVT to find a zero (i.e., a root) of a function requires having the following objects and proofs:

1. a continuous function $f(x)$ from reals to reals, which is defined at least on some closed interval;
2. two numbers $a$ and $b$ such that $a<b$ and $f(x)$ is defined on $[a,b]$;
3. a proof that $f(a)<0$ and $f(b)>0$, or that $f(a)>0$ and $f(b)<0$.

If you are using the IVT to prove that an equation $g(x)=h(x)$ has a solution, then take $f(x)=g(x)-h(x)$. In particular, if you need to find not a zero of a function, but a solution to $g(x)=c$ for some $c$, then you consider $f(x)=g(x)-c$. It is easy to see that $g(a)-c<0<g(b)-c\iff g(a)<c<g(b)$.

In what you wrote, I don't always see all these elements.

So as Evgeny mentioned, driving 40 Miles per Hour on the highway and 25 Mile per Hour in the city, at some point in time between the highway and the city, the driver has gone 30 Miles per Hour. That's the one that stuck in my head the most.

40 is interval (b), per say, and 25 is (a)

\(\displaystyle 25 < Speed < 40\)

What is the function here? Presumably, it returns speed, but what is its argument and domain? Starting from post #2, I used $a$ and $b$ as the ends of a segment in the domain of a function. Of course, variable names don't matter, but it is crucial to have a function that is defined on some segment $[a,b]$. In this case, such function could map time or distance to speed.

So to sum it all up, to prove that \(\displaystyle x = cos(x)\) has a solution using continuity, we must see if it is continuous or not by turning it into a function. From there we have to find out any discontinuities (which it doesn't) so we know that x has a solution.

Remember I advised you to express yourself precisely? An equation may have a solution, but $x$ is not an equation, so it can't have a solution. You also did not write what function you are talking about. I assume it's $f(x)=x-\cos x$. More importantly, just the fact that $f(x)$ is continuous does not guarantee that there is a solution of the equation $f(x)=0$.

Given a good starting point is x = 0, it would be (0,-1). Given another good input is if f(x) = 0, around (0.739,0)

If you had a precise value of $x$ such that $f(x)=0$, then you would not need to invoke the IVT to show that $f(x)=0$ has a solution. Besides, you need precise values of $a$ and $b$ from the checklist above, not approximate values.

and that would be my intervals, -1 and 0.

$-1$ and $0$ are not intervals, they are numbers. Also recall that the checklist includes an interval $[a,b]$ from the domain, not the codomain (or range) of $f$. It seems to me that $-1$ and $0$ that you are considering are values of the function, so they come from the codomain.

So I suggest looking at the graph of $f(x)=x-\cos x$ and coming up with all objects mentioned in the checklist above.

 

[Deveno]

As Evgeny points out, you need to have a clearer idea what the different "parts" of a function are.

Naively, there are 3 parts to a function:

1. The domain.

If we are thinking of a function as defined by pairs of points in the plane:

$(x,y)$, where $y = f(x)$

The domain is all the things $x$ could be. Why can't $x$ be "anything"? Well we can write lots of expressions that don't make sense, for example:

$\dfrac{2x}{x - x}$ is going to be hard to evaluate, no matter what $x$ is. So points $x$ for which $f(x)$ makes no sense are EXCLUDED from the domain.

2. The range. There is another thing called "the co-domain", but for now, let's just say that the co-domain (the set where $f(x)$ takes its values in) is always going to be $\Bbb R$.

The range is all possible things $y$ (which equals $f(x)$) could be. Sometimes it's a challenge to actually figure this out.

3. The "$f$" part, or the RULE, by which we determine what $y$ is, given $x$.

Now with continuous functions (and the precise definition of what "continuous" means will have to wait until you cover limits-for now, a continuous function is one you can draw without the pencil leaving the paper), we are often interested in what $f$ looks like when the domain is an INTERVAL (a continuous "segment" of the real line, usually drawn on "the $x$-axis").

The problem you are trying to solve is not like "solving a quadratic equation", you're not going to find out "what $x$ is" when $f(x) = 0$. You are doing another thing entirely: proving that some $x$ EXISTS, with $f(x) = 0$.

Unless you use fairly sophisticated methods, finding an "exact" value for such an $x$ is going to be rather difficult. However, we can get a general idea of "where" this $x$ is, by "trapping" it in an interval.

You are off to a good start with $f(x) = x - \cos x$ by looking at $f(0)$, which is $-1$. What is $f(\frac{\pi}{2})$?