function ?

[edieber]

how do I solve this one?
y=-2x^3-9x^2-60x

[Muzza]

You're going to have to supply more information about the problem. The actual question would be good.

[edieber]

You're going to have to supply more information about the problem. The actual question would be good.

this it function inquiry question?

[omicron]

I don't think that is gonna help.

[T@P]

you want to solve for the roots of this function?

in that case, your function is
0 = -2x^3 -9x^2 -60x solving for x
in that case, factor out x from the right hand side,
you get x \cdot (-2x^2 -9x -60) = 0
then you can use the quadratic formula to solve for x.
x would equalt to \frac{-b \pm \sqrt{b^2 -4ac}}{2a}
which by pluging and chuging gives you two answer.
The final answer would then be x = 0, r_{1}, r_{2}

in the case that you are attempting to simply draw it with the help of calculus, i do not see where the problem lies. Simply take the first derivative, set it to 0, and continue normally. perhaps a little more information on what your desired answer is...?

[marlon]

how do I solve this one?
y=-2x^3-9x^2-60x

Perhaps, i am missing something here, but what's all the fuss about???

Just write : -x(2x^2+9x+60) = 0. So 0 is one solution. Then you have to solve 2x^2+9x+60 = 0. Since D = 81-2*4*60 < 0 You have no real solutions. can you incorporated compelx numbers ? If so, just continue as in the D > 0 case. if not, there are no solutions here...


marlon

[T@P]

good pont marlon, there is only one solution. it is x = 0. sorry i didnt even bother to check what D was.

[Muzza]

Perhaps, i am missing something here, but what's all the fuss about???


Most of us are not mindreaders, and don't know what people are really asking when they say "solve" and then post a function.

[marlon]

Most of us are not mindreaders, and don't know what people are really asking when they say "solve" and then post a function.


Sorry, but do you know another way to solve a function ???

i think not...

marlon, the mind-reader

[Muzza]

I don't know of any way to "solve a function", just like I don't know how to "solve a banana".

[marlon]

mmmmm

sweden, ja???

are you into ABBA ???

marlon

[Muzza]

Nope. How many waffles have you eaten today...?

[marlon]

5

marlon

[dav2008]

Sorry, but do you know another way to solve a function ???

i think not...

marlon, the mind-reader
y=x

solve that function

[marlon]

y=x

solve that function


hahaha, please ask a more difficult question...


answer : x = 0

marlon

[marlon]

Solving means finding the solutions guys...

wow, what a revealing theory...

marlon

[dav2008]

hahaha, please ask a more difficult question...


answer : x = 0

marlonThat answer is only true if y is 0.

[marlon]

That answer is only true if y is 0.
yes indeed it is...

marlon

[marlon]

if y were to be 5 the the question should be : solve x - 5. hence the answer is 5...


beware that you do not violate the injectivity part of the definition of a function. With one x value, there can be at least one y-value, otherwise we do not have a function. Eg. x = 6 is NOT a function. y = x - 6 sure is...
Ofcourse with one y-value there can be several x-values like in y = x²


marlon

[HallsofIvy]

"Solving means finding the solutions guys...

wow, what a revealing theory...

marlon"

You still haven't understood what everyone is saying! Yes, "solve" means find the solutions- but to what problem?

The orginal post just said "How do I solve this one? y= 3x³ -9x²- 60x"

That's not a "problem" that's just a statement. You then ASSUMED that the problem was "find all values of x that make y= 0" but that certainly is NOT the only possible problem that could be associated with a function.

[marlon]

"Solving means finding the solutions guys...

wow, what a revealing theory...

marlon"

You still haven't understood what everyone is saying! Yes, "solve" means find the solutions- but to what problem?

The orginal post just said "How do I solve this one? y= 3x³ -9x²- 60x"

That's not a "problem" that's just a statement. You then ASSUMED that the problem was "find all values of x that make y= 0" but that certainly is NOT the only possible problem that could be associated with a function.

When no other specific y-values are given it sure is the ONLY possible problem. Let's stop the vagueness here, please...

regards
marlon

[matt grime]

marlon, they're having you here: one does not solve functions in the sense you think they mean. One solves equations, ie finds their roots, but functions don't possess roots in this sense.

[marlon]

marlon, they're having you here: one does not solve functions in the sense you think they mean. One solves equations, ie finds their roots, but functions don't possess roots in this sense.

hhhmmmm, this doesn't sound really convincing, but nevertheless...Though i get your point and I agree with you "in this sense" and disagree with you "in that sense", i can only conclude by saying you are making something real easy very difficult. Personally, i think you know very well what i mean. I asked before to stop this vagueness because it makes the original question sound more difficult then it is, yet the answer is correct.

regards
marlon

[HallsofIvy]

So what you are really telling us is that f(x)= 0 is the kind of problem YOU mostly see and you are making an assumption based on YOUR limited experience. Yes, we could have also jumped to that conclusion but most of us have seen many kinds of problems based on a polynomial. Asking for clarification is NOT "making it more difficult". YOU may be satisfied with "an" answer, whether it is right or wrong, we are not. The only person who can clarify the problem is edieber, the original poster, who apparently hasn't even bothered to read the responses to his/her question.

[Manchot]

When no other specific y-values are given it sure is the ONLY possible problem. Let's stop the vagueness here, please...

regards
marlonWhat about finding the extrema of the function, and/or sketching it? As HallsOfIvy said, there are quite a few possibilities for the meaning of "solve" in this case.

[marlon]

So what you are really telling us is that f(x)= 0 is the kind of problem YOU mostly see and you are making an assumption based on YOUR limited experience. Yes, we could have also jumped to that conclusion but most of us have seen many kinds of problems based on a polynomial. Asking for clarification is NOT "making it more difficult". YOU may be satisfied with "an" answer, whether it is right or wrong, we are not. The only person who can clarify the problem is edieber, the original poster, who apparently hasn't even bothered to read the responses to his/her question.

I most certainly disagree with this "personal"-post. Please, post only if you have anything interesting to say, you are not making any point here. I never said that asking for clarification equals making things more difficult, that is a very easy and useless statement. There are different types of questions concerning polynomials, i do know that so sorry that you are not the only one here. Yet in my opinion it was clear what was meant by the original post so let's not continue this useless discussion on personal views. The questions were asked and the answers were given...that's all there is to it...

marlon

[marlon]

What about finding the extrema of the function, and/or sketching it? As HallsOfIvy said, there are quite a few possibilities for the meaning of "solve" in this case.

Indeed, the post of Hallsofivy contained no new info...

Finding extrema is not the same as asking for a solution. In this case just ask for the extrema. Don't make things so vague, You see my point ??? If a sketch was asked, i am sure the original poster would have shown us the intelligence to formulate the question in such a way that the word "sketch" would have been in it...just my opinion...

marlon

[matt grime]

Marlon, you were the person who said that some philosophers were speaking of where they knew not and were misusing mathematics, right? Well, saying solve f(x) for some function is also a misuse of mathematical terms.

Whether or not, in your opinion (as you state it to be), the question was clear doesn't mean that the question was actually correct. There may be some element of playing devil's advocate going on, but it is better to stop people misusing terms than letting them carry on being wrong, surely?

[JonF]

Marlon you apparently have no idea what a function is. I see that you have read the formal definition, that is good. But, apparently it’s meaning was lost on you.

A function simply is some transformation on a thing (let’s call it a dependant variable) turning into another thing (let’s call it a dependant variable). The function also has another requirement, when you transform an independent variable you only get one result.

So when you ask me to solve

f(x)=-2x^3-9x^2-60x

With out telling me what you want me to transform that independent variable it into, you aren’t supplying sufficient information.

The “-2x^3-9x^2-60x” part is what the transformation this particular function is. It takes an input, cubes it then, multiplies it by –2. After that it takes that same input squares it and multiplies it by –9. Then it takes the input and multiplies it by –60. And lastly it takes those three values and adds them together. What you wanted to know is when will this process give a result of 0. Another equally valid question is when will this process give a result of 10? Or 20?

Functions are very different from equations. With an equation you are trying to find a solution. Functions are entirely different ideas. With functions you give an input and get an output.

For example the function of your height over time could be: for all 0<t<20: f(t) = t(t-20)^(1/2) where f(t) is in inches and t is in years. It makes no sense to solve this equation for 0. Why would you want to know when you were 0 inches tall? But you might want to find out when (or if) you were going to be 6 feet tall. Which would be 72= t(t-20)^(1/2)


Back to your function. What you wanted to ask is what independent variable will make the function yield a value of 0. I.e. solve f(x)=0

But let’s say you wanted to figure out when the function gave a value of, oh 20. I.e. f(x)=20

Then you would get: 20=-2x^3-9x^2-60x
0=-2x^3-9x^2-60x-20

[marlon]

Marlon, you were the person who said that some philosophers were speaking of where they knew not and were misusing mathematics, right? Well, saying solve f(x) for some function is also a misuse of mathematical terms.

Whether or not, in your opinion (as you state it to be), the question was clear doesn't mean that the question was actually correct. There may be some element of playing devil's advocate going on, but it is better to stop people misusing terms than letting them carry on being wrong, surely?

Hi matt,
the way i see it is that the question "solve f(x)" means solving for f(x) = 0. If another number is specified then the question is ofcourse cristal clear yet if no number is specified the f(x) = 0 is meant since one must interprete "to solve" as "find the solutions of the equation f(x) = 0". But ofcourse one can argue this just like one can argue anything else, so i suggest we let this subject drop and move on to solving REAL problems...

best of regards
marlon

[marlon]

Marlon you apparently have no idea what a function is. I see that you have read the formal definition, that is good. But, apparently it’s meaning was lost on you.

A function simply is some transformation on a thing (let’s call it a dependant variable) turning into another thing (let’s call it a dependant variable). The function also has another requirement, when you transform an independent variable you only get one result.

So when you ask me to solve

f(x)=-2x^3-9x^2-60x

With out telling me what you want me to transform that independent variable it into, you aren’t supplying sufficient information.

The “-2x^3-9x^2-60x” part is what the transformation this particular function is. It takes an input, cubes it then, multiplies it by –2. After that it takes that same input squares it and multiplies it by –9. Then it takes the input and multiplies it by –60. And lastly it takes those three values and adds them together. What you wanted to know is when will this process give a result of 0. Another equally valid question is when will this process give a result of 10? Or 20?

Functions are very different from equations. With an equation you are trying to find a solution. Functions are entirely different ideas. With functions you give an input and get an output.

For example the function of your height over time could be: for all 0<t<20: f(t) = t(t-20)^(1/2) where f(t) is in inches and t is in years. It makes no sense to solve this equation for 0. Why would you want to know when you were 0 inches tall? But you might want to find out when (or if) you were going to be 6 feet tall. Which would be 72= t(t-20)^(1/2)


Back to your function. What you wanted to ask is what independent variable will make the function yield a value of 0. I.e. solve f(x)=0

But let’s say you wanted to figure out when the function gave a value of, oh 20. I.e. f(x)=20

Then you would get: 20=-2x^3-9x^2-60x
0=-2x^3-9x^2-60x-20

really??? :rolleyes: :rolleyes:



ps : make sure that if you wanna correct someone, you do it the right way. Your definition of a function is not complete. For example can x = 6 be catagorized as a real function conform is mathematical definition??? Besides a function is a relation and not a transformation.

Obviously you need to be more correct in your corrections :wink: :wink:

marlon