Functions

[Manalzbair]

We have :
f(x) =(1-x) ÷x, x is in IR*
g(x)=|f(x)|
h(x) =4-x^2
Solve g(x)-h(x) <0 (not with a graph)
( I tried solving so , S=]a, b[ U ]c,d[., but i don't know what are a, b, c and d. I would appreciate the help a bunch)

 

[Dr.Peterson]

We have :
f(x) =(1-x) ÷x, x is in IR*
g(x)=|f(x)|
h(x) =4-x^2
Solve g(x)-h(x) <0 (not with a graph)
( I tried solving so , S=]a, b[ U ]c,d[., but i don't know what are a, b, c and d. I would appreciate the help a bunch)

Please show what you've tried, not just the form of the answer you are expecting. Also, what topics have you recently learned?

We need to have some idea of what methods you can handle, and whether you did something reasonable but made a little mistake.

Also, does "IR*" mean the non-zero reals?

 

[Manalzbair]

Please show what you've tried, not just the form of the answer you are expecting. Also, what topics have you recently learned?

We need to have some idea of what methods you can handle, and whether you did something reasonable but made a little mistake.

Also, does "IR*" mean the non-zero reals?

Logic, applications of sets, and generality of functions. I solved so with a graph, but the solution won't be exact.
And yes it does, I'm new here still don't know how to use the site properly.

 

[Dr.Peterson]

Your description of what you have learned is very vague. Have you learned about solving polynomial or rational inequalities?

I hoped to see some attempt at solving the inequality algebraically! The graph, which is good, helps to see what to expect, but not to give the actual solution (unless the values turned out to be simple points you'd graphed).

The inequality, after replacing functions with their expressions, turns out to be

|(1 - x)/x| - (4 - x^2) < 0​

To handle an absolute value, I would start by making cases: Find when (1 - x)/x is positive and negative, and in the latter case, the absolute value will equal -(1 - x)/x. (The graph sketch is helpful here.)

This gives you two different rational inequalities to solve within different regions. The standard way to solve these is to combine them into a single fraction (using a common denominator) and factor to find when the numerator and denominator are zero. Does any of that sound familiar?

So, what can you do with the first case"

(1 - x)/x - (4 - x^2) < 0​

 

[Manalzbair]

Your description of what you have learned is very vague. Have you learned about solving polynomial or rational inequalities?

I hoped to see some attempt at solving the inequality algebraically! The graph, which is good, helps to see what to expect, but not to give the actual solution (unless the values turned out to be simple points you'd graphed).

The inequality, after replacing functions with their expressions, turns out to be

|(1 - x)/x| - (4 - x^2) < 0​

To handle an absolute value, I would start by making cases: Find when (1 - x)/x is positive and negative, and in the latter case, the absolute value will equal -(1 - x)/x. (The graph sketch is helpful here.)

This gives you two different rational inequalities to solve within different regions. The standard way to solve these is to combine them into a single fraction (using a common denominator) and factor to find when the numerator and denominator are zero. Does any of that sound familiar?

So, what can you do with the first case"

(1 - x)/x - (4 - x^2) < 0​

This is what i did so far, and i didn't understand how to combine them is like this : 1/x (2x^3 - 8x)<0?

 

[Dr.Peterson]

This is what i did so far, and i didn't understand how to combine them is like this : 1/x (2x^3 - 8x)<0?

Where did that come from? You had it right, here:





You don't combine these; you solve them one at a time, each within its own interval according to the case it represents. The first will provide three of your endpoints, and the second the other.

But I'm not sure what you are expected to do next, as the cubics are not factorable. Are you allowed to use some sort of technology?

 

[Manalzbair]

Where did that come from? You had it right, here:

View attachment 30003

View attachment 30002

You don't combine these; you solve them one at a time, each within its own interval according to the case it represents. The first will provide three of your endpoints, and the second the other.

But I'm not sure what you are expected to do next, as the cubics are not factorable. Are you allowed to use some sort of technology?

We haven't studied 3rd degree inequalities yet. But I'll try to understand it and solve it. Can you explain how the first one provides 3 end points and the second one the other point? And thanks a lot for yr help and patience Dr. Peterson

 

[Dr.Peterson]

We haven't studied 3rd degree inequalities yet. But I'll try to understand it and solve it. Can you explain how the first one provides 3 end points and the second one the other point? And thanks a lot for yr help and patience Dr. Peterson

To repeat, I don't know a method you would have been taught for finding the zeros of these cubics! You should hopefully have learned about rational inequalities in general, and be able to solve them when you can factor the polynomials; but these can't be easily factored. I have no idea what you are expected to do; you haven't answered my question about whether you are allowed to use some form of technology such as a graphing calculator or computer program. That is the only way I can solve it. Have you asked your teacher yet?

Each of the two cases restricts the domain to certain values of x; it happens that three of the zeros are in one domain, and one is in the other.

 

[Manalzbair]

To repeat, I don't know a method you would have been taught for finding the zeros of these cubics! You should hopefully have learned about rational inequalities in general, and be able to solve them when you can factor the polynomials; but these can't be easily factored. I have no idea what you are expected to do; you haven't answered my question about whether you are allowed to use some form of technology such as a graphing calculator or computer program. That is the only way I can solve it. Have you asked your teacher yet?

Each of the two cases restricts the domain to certain values of x; it happens that three of the zeros are in one domain, and one is in the other.

My teacher wouldn't give us the answer and the explanation, he asked us to try solving it and then he didn't talk about it. I'm allowed to use anything as long as its familiar to what I've studied before, which is like solving 2nd degree inequalities, generality of functions (just the basic functions)

 

[JasonTurner]

Before starting any math problem, it is ideal for determining what minute details we have. Jumping straight to the solution makes us leave the basics. In your question, we have a couple of key arrangements and tips. Firstly, you have the term IR in the first sentence. This is known as Identity Function. Identity Function states that while we f(x) and g(x), then g(x) is the exact inverse of f(x). In essence, if you solve for f(x), you will have g(x) that will inverse it. However, this inversion is not in magnitude. This inversion is in the sign.

Next, you have another term of h(x), and all of a sudden, you begin to lose confidence. Three terms in a problem do not require a combined calculation. For example, like we have two equations to solve simultaneously, there is no such combination of three. The third equation always relates to any of the two. Thus, in your question, the complete answer you need without a graph is between g(x) and h(x). However, the question demands that you solve for f(x) first and then use the IR condition and have g(x). As of a, b, c, and d, I think these are the coordinates from the graph. But considering your questions, it is hard to relate a, b, c, and d with the equations.

 

[Dr.Peterson]

Before starting any math problem, it is ideal for determining what minute details we have. Jumping straight to the solution makes us leave the basics. In your question, we have a couple of key arrangements and tips. Firstly, you have the term IR in the first sentence. This is known as Identity Function. Identity Function states that while we f(x) and g(x), then g(x) is the exact inverse of f(x). In essence, if you solve for f(x), you will have g(x) that will inverse it. However, this inversion is not in magnitude. This inversion is in the sign.

Next, you have another term of h(x), and all of a sudden, you begin to lose confidence. Three terms in a problem do not require a combined calculation. For example, like we have two equations to solve simultaneously, there is no such combination of three. The third equation always relates to any of the two. Thus, in your question, the complete answer you need without a graph is between g(x) and h(x). However, the question demands that you solve for f(x) first and then use the IR condition and have g(x). As of a, b, c, and d, I think these are the coordinates from the graph. But considering your questions, it is hard to relate a, b, c, and d with the equations.

I don't think anything you've said here makes sense. What I marked in bold is simply false.

What is your goal here?

 

[JasonTurner]

I don't think anything you've said here makes sense. What I marked in bold is simply false.

What is your goal here?

You can say it is completely false because you haven’t looked for what IR means in general. It represents Identity Function in expansion. And the rule for Identity Function states that you have two terms of f(x) and g(x). Moreover, you will have to solve for f(x) in the first stage. Once you solve it, the answer to g(x) will be of the same magnitude but opposite in sign. In other words, the answer will be the same in magnitude but opposite in direction. Everything that you have highlighted will become non-highlighted if you consult different sources!

 

[Dr.Peterson]

You can say it is completely false because you haven’t looked for what IR means in general. It represents Identity Function in expansion.

It's true that the notation IR (actually, [imath]I_{\mathbb{R}}[/imath]) can be used for the identity function (namely [imath]f(x) = x[/imath]) on the set of real numbers, [imath]\mathbb{R}[/imath].

But meaning depends on context; words and symbols don't mean the same thing everywhere they are used.

The context here is "f(x) =(1-x) ÷x, x is in IR*", which makes it clear that "IR*" is being used to name a set, not a function; a variable can't be "in" a function. My assumption was that the OP meant [imath]\mathbb{R}^*[/imath], the set of non-zero real numbers, and the OP agreed (post 3). It is not good practice to represent "blackboard bold" R as "IR", but I have seen it done enough to understand it.