Word problem. Don't understand it. Please explain what's asked of me!

[Kamhogo]

"A continuous function f (x) is positive at x=0 and negative for x=1. How many solutions does f(x)=0 have between x=0 ans x=1? Explain. " I don't have any work to show since I don't even understand what's asked of me. Please help.

 

[ksdhart]

First off, I'd note that since the function is continuous (presumably everywhere since the problem doesn't specify), that means you can draw its graph without having to pick up your pencil. In this case, only the portion of the graph between x = 0 and x = 1 is relevant. So, pick two points, one where x = 0 at arbitrary positive y-value, and the other at x = 1, with some negative y-value. Draw any continuous function you like connecting those two points. What do you notice? Specifically, how many times did you cross the x-axis? So, how many points does f(x) = 0 in that interval? Now start again and draw a different function connecting the two points. What do you notice now? What, then, do you think the answer to the word problem is?

 

[Kamhogo]

First I connected the two points in a straight line and it seemed like there was only one solution for f (x)=0. But then I connected the points in a sinusoidal way and it seems like there is an infinity of solutions (I opt for this answer).

 

[stapel]

First I connected the two points in a straight line and it seemed like there was only one solution for f (x)=0. But then I connected the points in a sinusoidal way and it seems like there is an infinity of solutions (I opt for this answer).

So the number of solutions (x-intercepts, roots, etc) depends on what sort of curve f(x) is. But they haven't specified what sort of curve it is. Either way, though, I'll bet you get say something about the minimum number of solutions on the interval, right? "f(x) has at least...."?

 

[Kamhogo]

☺ f(x)=0 has at least one solution. Since the function is continuous, the line/ curve has to cross the x-axis at least once...Correct?
Question: what do you mean by roots etc?

 

[Bob Brown MSEE]

also

You can also say whether ther is an even or odd number of solutions.

 

[Kamhogo]

Bob, would you give me a hint as how to determine that please?

 

[ksdhart]

Question: what do you mean by roots etc?

In the context of a function or a polynomial, there are a lot of words you can use to indicate the same thing. Some teachers/textbooks might say "Solve this polynomial," which really means find all the values the function is equal to 0. Similarly, you might be asked to "Find any roots of this polynomial" or "Find any x-intercepts of this polynomial" and both mean the same. That was all that Stapel was indicating there.

Bob, would you give me a hint as how to determine [if there is an odd or even number of solutions] please?

I think this is a actually a really good question. I can explain it, but I personally think it's a little hand-wavey and leaves me feeling "Ugh." The basic idea is to consider the function over the interval [0,1]. It's continuous and you start at a positive and go to a negative, so we've correctly established you must cross the x-axis at least once. Now, consider if you crossed the x-axis twice. You'd go from positive to negative back to positive... oops. However, there's an important part missing from that consideration. For the function to have a root, it must be equal to 0, but that doesn't actually mean it has to cross the x-axis. What if your function swings down to negative y-value comes back up and touches the x-axis but doesn't cross into positives?

As an example, look at the function $\displaystyle x^3+2x^2-7x+4$. It has two x-intercepts (at x = -4 and x = 1), so I'd be inclined to say that it has two roots. However, that's not quite right. The fundamental theorem of algebra says that every nth-degree polynomial has exactly n solutions. Some of them might be complex, and others real. Even more confusingly, some of the roots might be the same real number. In this example polynomial, the roots are -4, 1, and... 1. Now you can perhaps see why I say this concept leaves me unsettled.

In general, if the graph of the polynomial crosses the x-axis, that root has odd multiplicity, meaning there are an odd number of roots at that x-value. If the graph "bounces off" (i.e. it touches the x-axis but never the value never changes signs), then that root had even multiplicity and there are an even number of roots at that x-value. Given this knowledge, what can you say about whether your hypothetical function f(x) has an even or an odd number of roots?

 

[Dale10101]

On the road

"A continuous function f (x) is positive at x=0 and negative for x=1. How many solutions does f(x)=0 have between x=0 ans x=1? Explain. " I don't have any work to show since I don't even understand what's asked of me. Please help.

If you are on the positive side of the road (x-axis) and wish to cross to the negative side of the road you must cross the road at least once, if you cross back for two crossings of the road then you must cross back for a third crossing of the road to return to the negative side of the road. You are allowed to cross the road as many times as you wish (strictly keeping your feet on the ground) but in the end to get to the negative side you will have needed to cross the road an odd number of times ... for a pedestrian but hopefully useful picture.

Suppose you wanted to cross the road once, at, say x = 1/3

Then (x-1/3)= 0 is the polynomial equation you would write as the statement evaluating the polynomial function
f(x) = 1/3 at x = 0.

Suppose you wanted to cross the road twice at, say x = 1/3 and x = 4/7.

Then (x-1/3)(x-4/7) = 0 is the polynomial equation you would write as the statement evaluating the polynomial function

f(x) =(x-1/3)(x-4/7) = 21x² -19x +4 = 0, at x = 0.

This double crossing of the road would put you back on the positive side of the road so you might want to cross again.

Suppose then you wanted to cross the road 3 times at say x = 1/3, 4/7 and 8/9. Then

Then (x-1/3)(x-4/7)(x-8/9) = 0 is the polynomial equation you would write as the statement evaluating the polynomial function

f(x) = (x-1/3)(x-4/7)(x-8/9) = x^3 - 113x^2/63 + 188x/189 - 32/189 = 0, at x = 0.

These examples illustrate

A) The general idea of the somewhat amazing fact that a polynomial expressed as a sum of monomials can also be expressed as a product of factors (and vice versa). An amazement akin to the fact that a natural number can always be expressed as the product of prime numbers.

B) What is meant by a "root" or "zero" of a function

C) Also the apparent equivalence of x's that correspond to an f(x) = 0 with a crossing of the x - axis. HOWEVER ....

as Ksdhart pointed out, any such point, i.e. x = 1/3, 4/7, 8/9 might in fact correspond not to a crossing of the x-axis but rather a "touching" of the x-axis, so you would want to distinguish the cases by either graphing the function you created, or using a calculus procedure to establish that such a particular point is not a local minimum or maximum value.

D) In general, the degree of the univariate polynomial corresponding to it's number of zero's (touches or crossings of the x- axis).

Additional point: I am not sure that an actual answer has been expressed. At first blush, many, including myself, were equating the correspondence of a root , a value of x that corresponds to f(x) = 0 with a crossing of the x-axis, the usual case. It would seem that in fact one can only say that the number of solutions is "one, or more" since even one root that corresponds to a touch rather then a crossing of the x-axis would prevent one from saying that an odd number of roots are required to end up below the x-axis in this case. For example if the first root's corresponding f(x) = 0 crossed from above to below the x-axis and the second root's f(x) = 0 merely touched the x-axis and retreated towards negative infinity then you would have an f(x) with an even number of roots that crossed the x-axis one time.

 

[Ishuda]

......
A) The general idea of the somewhat amazing fact that a polynomial expressed as a sum of monomials can also be expressed as a product of factors (and vice versa). An amazement akin to the fact that a natural number can always be expressed as the product of prime numbers.
...

Couldn't resist: Any continuously differentiable function f(x) which has a zero at x=x₁ can be expressed as
f(x) = (x-x₁) g(x)
where g(x) is continuous. What is g(x) you ask? Well obviously it is
g(x) = $\displaystyle \frac{f(x)}{x-x_1};\, x\, \ne\, x_1$
g(x) = $\displaystyle \frac{df(x)}{dx}|_{x=x_1};\, x\, =\, x_1$

From that, you can derive the above and the more general statement: If f(x) has n roots of multiplicity m_j, j=1, 2, 3, ...,n and M=max{m_j, j=1, 2, 3, ...,n} and if all derivatives of f(x) through order M are continuous, then f(x) can be written as
f(x) = $\displaystyle [\, \Pi_{j=1}^{j=n}\, (x-x_j)^{m_j}\, ]\, g(x)$
where g(x) is continuous [in the case of a polynomial g(x) would be a constant].

 

[Kamhogo]

Wow. I'm blown away. I've been reflecting on all that (especially when I cross the road ?) and I just love math even more. Thanks for the explanations everyone!

 

[Dale10101]

Question

Couldn't resist: Any continuously differentiable function f(x) which has a zero at x=x₁ can be expressed as
f(x) = (x-x₁) g(x)
where g(x) is continuous. What is g(x) you ask? Well obviously it is
g(x) = $\displaystyle \frac{f(x)}{x-x_1};\, x\, \ne\, x_1$
g(x) = $\displaystyle \frac{df(x)}{dx}|_{x=x_1};\, x\, =\, x_1$

From that, you can derive the above and the more general statement: If f(x) has n roots of multiplicity m_j, j=1, 2, 3, ...,n and M=max{m_j, j=1, 2, 3, ...,n} and if all derivatives of f(x) through order M are continuous, then f(x) can be written as
f(x) = $\displaystyle [\, \Pi_{j=1}^{j=n}\, (x-x_j)^{m_j}\, ]\, g(x)$
where g(x) is continuous [in the case of a polynomial g(x) would be a constant].

I would really like to understand what you are saying.

I am not entirely lost (I think) but one clue please. In your second expression for g(x), does it read that g(x) is equal to the derivative of f(x) with respect to dx but only when x = x₁, that is, the domain for g(x) is effectively reduced to the single value x₁? It seems to me at this point that you are describing a differential, that is, the slope of f(x) at x = x₁ rather then the derivative of f(x). Basically can you tell me what this second statement for g(x) is describing.

I can see how taking the limit of the right side of the first statement for g(x) is related to the definition of a derivative but am having this preliminary difficulty of interpreting the result. Thank you.

"From that, you can derive the above …."

My work so far:

The first step is to derive eq 2 from eq 1

$${\rm{eq1: }}g(x) = \frac{{f(x)}}{{(x - {x_1})}}$$

$${\rm{eq2: }}g(x) = \frac{{df(x)}}{{dx}}{|_{x = {x_1}}};\,\,\,\,x = {x_1}$$

\[\begin{array}{l}
{\rm{eq3: }}g(x) = \frac{{f(x) - f({x_1})}}{{(x - {x_1})}}\,\,,\,\,\,\,\,{\rm{since }}f({x_1}) = 0\\
{\rm{by the definition of a derivative,}}\\
{\rm{eq4: }}\frac{{df(x)}}{{dx}} = {\rm{ }}\mathop {\lim }\limits_{x \to {x_1}} \,\,\frac{{f(x) - f({x_1})}}{{(x - {x_1})}}\\
eq4a:\,\,\,\,\mathop {\lim }\limits_{x \to {x_1}} \,\,g(x) = {\rm{ }}\,\mathop {\lim }\limits_{x \to {x_1}} \,\,\frac{{f(x) - f({x_1})}}{{(x - {x_1})}},\,\,\,\,\,\,\,\,\,\,f({x_1}) = 0,\,\,\,\,\,\,{\rm{so that,}}\\
{\rm{eq5: }}\mathop {\lim }\limits_{x \to {x_1}} \,\,g(x) = \,\frac{{d(f(x))}}{{dx}}{|_{x = {x_1}}}{\rm{, and so}}\\
{\rm{eq6: f(x) = }}\left[ {{\rm{(x - }}{{\rm{x}}_1})\,\frac{{d(f(x))}}{{dx}}} \right]{|_{x = {x_1}}}
\end{array}\]

 

[Ishuda]

I would really like to understand what you are saying.

I am not entirely lost (I think) but one clue please. In your second expression for g(x), does it read that g(x) is equal to the derivative of f(x) with respect to dx but only when x = x₁, that is, the domain for g(x) is effectively reduced to the single value x₁? It seems to me at this point that you are describing a differential, that is, the slope of f(x) at x = x₁ rather then the derivative of f(x). Basically can you tell me what this second statement for g(x) is describing.

I can see how taking the limit of the right side of the first statement for g(x) is related to the definition of a derivative but am having this preliminary difficulty of interpreting the result. Thank you.

"From that, you can derive the above …."

My work so far:

The first step is to derive eq 2 from eq 1

$${\rm{eq1: }}g(x) = \frac{{f(x)}}{{(x - {x_1})}}$$

$${\rm{eq2: }}g(x) = \frac{{df(x)}}{{dx}}{|_{x = {x_1}}};\,\,\,\,x = {x_1}$$

\[\begin{array}{l}
{\rm{eq3: }}g(x) = \frac{{f(x) - f({x_1})}}{{(x - {x_1})}}\,\,,\,\,\,\,\,{\rm{since }}f({x_1}) = 0\\
{\rm{by the definition of a derivative,}}\\
{\rm{eq4: }}\frac{{df(x)}}{{dx}} = {\rm{ }}\mathop {\lim }\limits_{x \to {x_1}} \,\,\frac{{f(x) - f({x_1})}}{{(x - {x_1})}}\\
eq4a:\,\,\,\,\mathop {\lim }\limits_{x \to {x_1}} \,\,g(x) = {\rm{ }}\,\mathop {\lim }\limits_{x \to {x_1}} \,\,\frac{{f(x) - f({x_1})}}{{(x - {x_1})}},\,\,\,\,\,\,\,\,\,\,f({x_1}) = 0,\,\,\,\,\,\,{\rm{so that,}}\\
{\rm{eq5: }}\mathop {\lim }\limits_{x \to {x_1}} \,\,g(x) = \,\frac{{d(f(x))}}{{dx}}{|_{x = {x_1}}}{\rm{, and so}}\\
{\rm{eq6: f(x) = }}\left[ {{\rm{(x - }}{{\rm{x}}_1})\,\frac{{d(f(x))}}{{dx}}} \right]{|_{x = {x_1}}}
\end{array}\]

Yes, except for Eq. 4 & Eq. 6. Those equations apply only at x=x₁ , that is for f(x₁) and f'(x₁), not f(x) nor f'(x)

That is, since f is differential, its derivative is continuous, and f(x₁)=0, the function
g(x) = $\displaystyle \frac{f(x)}{x\, -\, x_1}$
has a removable singularity at x=x₁.