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Posted: 1/29/2006 2:42:10 PM EDT
I just do not understand what the hell this question is asking! Check this out:

"For f(x) = x3 + 2x2 – 5x + 8, find f(-1)."

Does that mean substitute negative 1 for the x variable? If so, WTF do I do with 2x2? 2[-1(2)]? Is this asking for a solution, a simplification?

Somebody smarter than me (like that's a task) tell me what the hell this gibbberish means?
Link Posted: 1/29/2006 2:48:26 PM EDT
[Last Edit: 1/29/2006 2:52:27 PM EDT by jeep44]
14

I read it as: 3x + (2*2)-5x +8


Ooh! ooh! I didn't even think of exponents....
Link Posted: 1/29/2006 2:50:32 PM EDT
[Last Edit: 1/29/2006 2:53:45 PM EDT by H46Driver]
Are the numbers after the x exponents? If so, the answer is 14, but not computed as posted above.

(-1)(-1)(-1) + 2(-1)(-1) -5(-1) + 8 =

-1 + 2 +5 +8 = 14
Link Posted: 1/29/2006 2:51:27 PM EDT
87.
Link Posted: 1/29/2006 2:51:58 PM EDT
[Last Edit: 1/29/2006 2:55:06 PM EDT by All_Beef_Patty]

Originally Posted By richardh247:
I just do not understand what the hell this question is asking! Check this out:

"For f(x) = x3 + 2x2 – 5x + 8, find f(-1)."

Does that mean substitute negative 1 for the x variable? If so, WTF do I do with 2x2? 2[-1(2)]? Is this asking for a solution, a simplification?

Somebody smarter than me (like that's a task) tell me what the hell this gibbberish means?



I think the exponents got screwed up. It looks like the equation should read:

f(x) = x^3 + 2x^2 - 5x + 8

In which case, f(-1) = -1 + 2 + 5 + 8 = 14.

Edited because I can't add.
Link Posted: 1/29/2006 2:53:31 PM EDT

Originally Posted By jeep44:
14

I read it as: 3x + (2*2)-5x +8



Exactly f(-1) = -1*3 + 2*2 - 5*-1 + 8

I forget the acronym for math operations order in an equation, but multiplication is before addition and subtraction.
Link Posted: 1/29/2006 2:54:40 PM EDT
[Last Edit: 1/29/2006 2:55:29 PM EDT by H46Driver]

Originally Posted By All_Beef_Patty:


In which case, f(-1) = -1 + 2 + 5 + 8 = 16.



You may want to check your addition ABP.

Unless CA is teaching that 15-1 = 16. Maybe that's how Arnold will balance the budget.
Link Posted: 1/29/2006 2:57:36 PM EDT
The equation reads:

'a function of x is equal to {(x to the third power) + (2 times x squared) minus (5x) plus 8} .

Evaluate the equation for a value of f at -1.

Not very difficult to solve.

Charles.
Link Posted: 1/29/2006 3:04:09 PM EDT

Originally Posted By Brian_in_Pullman:

Originally Posted By jeep44:
14

I read it as: 3x + (2*2)-5x +8



Exactly f(-1) = -1*3 + 2*2 - 5*-1 + 8

I forget the acronym for math operations order in an equation, but multiplication is before addition and subtraction.



PEMDAS

Parentheses
E­xponents
Multiplication
Division
Add­ition
Subtraction
Link Posted: 1/29/2006 3:08:23 PM EDT
I know the order of operations. I posted it EXACTLY as the question was written. I didn't change a thing, which is why I got so confused.
Link Posted: 1/29/2006 3:13:50 PM EDT
[Last Edit: 1/29/2006 3:15:20 PM EDT by richardh247]

Originally Posted By 1911builder:
The equation reads:

'a function of x is equal to {(x to the third power) + (2 times x squared) minus (5x) plus 8} .

Evaluate the equation for a value of f at -1.

Not very difficult to solve.

Charles.



Huh?

Edit: is (f) equal to (-1) or is (x)? There are no exponents in the problem, at least not how we usually do them like: 5^3 (five to the power of three).
Link Posted: 1/29/2006 3:41:47 PM EDT
C'Mon y'all, explain this to me in stupid talk.
Link Posted: 1/29/2006 3:45:30 PM EDT
[Last Edit: 1/29/2006 3:46:00 PM EDT by david_g17]

Originally Posted By richardh247:

Originally Posted By 1911builder:
The equation reads:

'a function of x is equal to {(x to the third power) + (2 times x squared) minus (5x) plus 8} .

Evaluate the equation for a value of f at -1.

Not very difficult to solve.

Charles.



Huh?

Edit: is (f) equal to (-1) or is (x)? There are no exponents in the problem, at least not how we usually do them like: 5^3 (five to the power of three).



from my experience, that equation is way too conicidental to not be exponents.
Link Posted: 1/29/2006 3:45:40 PM EDT

Originally Posted By richardh247:
C'Mon y'all, explain this to me in stupid talk.



f(x) = x^3 + 2x^2 - 5x + 8
= (x*x*x) + (2*x*x) - (5*x) + 8

so

f(-1) = (-1 * -1 * -1) + (2 * -1 * -1) - (5 * -1) + 8
= -1 + 2 + 5 + 8
= 14
Link Posted: 1/29/2006 4:54:49 PM EDT

Originally Posted By richardh247:
I just do not understand what the hell this question is asking! Check this out:

"For f(x) = x3 + 2x2 – 5x + 8, find f(-1)."

Does that mean substitute negative 1 for the x variable? If so, WTF do I do with 2x2? 2[-1(2)]? Is this asking for a solution, a simplification?

Somebody smarter than me (like that's a task) tell me what the hell this gibbberish means?



If that equation is indeed a precise rendition, then just substitute -1 for x at each location in the equation -

f(x) = f(x= -1) = f(-1) = (-1)3 + 2(-1)2 - 5(1-) + 8 = -3 -4 +5 +8 = 6

Actually, that is a good way to express that equation when testing student's knowledge about manipulating basic operations, but it is also a non-standard way of organizing the terms.
Link Posted: 1/29/2006 4:58:21 PM EDT

Originally Posted By richardh247:
I just do not understand what the hell this question is asking! Check this out:

"For f(x) = x3 + 2x2 – 5x + 8, find f(-1)."

Does that mean substitute negative 1 for the x variable? If so, WTF do I do with 2x2? 2[-1(2)]? Is this asking for a solution, a simplification?

Somebody smarter than me (like that's a task) tell me what the hell this gibbberish means?



X = -1

Substitute -1 for all of the Xs.

f(-1) = -3 + 2(-1)2 + 5 + 8

-7 + 5 + 8

-2 + 8

f(-1)= 6

f = -6 ?



Link Posted: 1/29/2006 5:00:27 PM EDT
[Last Edit: 1/29/2006 5:04:10 PM EDT by danpass]

Originally Posted By All_Beef_Patty:

Originally Posted By richardh247:
C'Mon y'all, explain this to me in stupid talk.



f(x) = x^3 + 2x^2 - 5x + 8
= (x*x*x) + (2*x*x) - (5*x) + 8

so

f(-1) = (-1 * -1 * -1) + (2 * -1 * -1) - (5 * -1) + 8
= -1 + 2 + 5 + 8
= 14




the formula 'translation' and the answer is correct
Link Posted: 1/30/2006 9:26:01 AM EDT

Originally Posted By danpass:

Originally Posted By All_Beef_Patty:

Originally Posted By richardh247:
C'Mon y'all, explain this to me in stupid talk.



f(x) = x^3 + 2x^2 - 5x + 8
= (x*x*x) + (2*x*x) - (5*x) + 8

so

f(-1) = (-1 * -1 * -1) + (2 * -1 * -1) - (5 * -1) + 8
= -1 + 2 + 5 + 8
= 14





richardh247 already stated that there are no exponents in the equation, and it's obvious from that post that he understands what that means.

Just because everyone's instinct is to rewrite the equation to conform with convention does not mean that it is in his problem, or even that it has to be. It's possible that formatting was lost somewhere along the way - maybe not. Consider the possiblilty that an exception occurred.


the formula 'translation' and the answer is correct

Link Posted: 1/30/2006 9:28:19 AM EDT
Don't you have a calculator?

Just press Y= and type in the equation man.
Link Posted: 1/30/2006 9:30:06 AM EDT

Originally Posted By AeroE:

Originally Posted By danpass:

Originally Posted By All_Beef_Patty:

Originally Posted By richardh247:
C'Mon y'all, explain this to me in stupid talk.



f(x) = x^3 + 2x^2 - 5x + 8
= (x*x*x) + (2*x*x) - (5*x) + 8

so

f(-1) = (-1 * -1 * -1) + (2 * -1 * -1) - (5 * -1) + 8
= -1 + 2 + 5 + 8
= 14





richardh247 already stated that there are no exponents in the equation, and it's obvious from that post that he understands what that means.

Just because everyone's instinct is to rewrite the equation to conform with convention does not mean that it is in his problem, or even that it has to be. It's possible that formatting was lost somewhere along the way - maybe not. Consider the possiblilty that an exception occurred.


the formula 'translation' and the answer is correct




its just weird that it shows x3 instead of 3x, which is why I also believe its trying to say x^3
Link Posted: 1/30/2006 10:47:40 AM EDT

Originally Posted By Tim84K10:
Don't you have a calculator?

Just press Y= and type in the equation man.



I have a TI BA II Plus, and it has no such button. Besides though, I don't want the calc to do the work. I want to know it - tht's the idea of college: to learn and then pass, not just to pass and walk away without retaining anything. If I can use the calc to check my answers, that would be great! But alas, I don't know what the "Y=" function does, nor does it appear I have one sigh.
Link Posted: 1/30/2006 10:59:18 AM EDT

Originally Posted By AeroE:

Originally Posted By danpass:

Originally Posted By All_Beef_Patty:

Originally Posted By richardh247:
C'Mon y'all, explain this to me in stupid talk.



f(x) = x^3 + 2x^2 - 5x + 8
= (x*x*x) + (2*x*x) - (5*x) + 8

so

f(-1) = (-1 * -1 * -1) + (2 * -1 * -1) - (5 * -1) + 8
= -1 + 2 + 5 + 8
= 14





richardh247 already stated that there are no exponents in the equation, and it's obvious from that post that he understands what that means.

Just because everyone's instinct is to rewrite the equation to conform with convention does not mean that it is in his problem, or even that it has to be. It's possible that formatting was lost somewhere along the way - maybe not. Consider the possiblilty that an exception occurred.


the formula 'translation' and the answer is correct




Does it make sense any other way?

We use ^ to denote exponents, so X to the third power is SUPPOSED to be written as X^3. That's where I am getting confused.

Two variables (xy) mean "x times y" but when we use numers it means "real number times variable."

So, 2x2 COULD mean "two to the power of two times two," "two times X to the power of two," or "two times X times two."

With the (-1) representing X it doesn't matter. But what is f(-3)? The the way the problem is solved would mean everything, and I just want to know if anyone has had a problem similar to this that could possibly tell me how I am to decipher a real number with a variable with a real number grouped together.

2(3^2) is different than {2[3(2)]} and so is 2[3(2)]^2.

See why I am confused? -1 won't change the solution, but if "f(-3) = f(x)" then we have a problem with 2x2. I'm not wanting an answer per se, I am wanting to know what in algebra dictates the way we approach such a thing so that I know in the future.

I really do appreciate everyone's help! Just trying to sort it all out.
Link Posted: 1/30/2006 11:10:04 AM EDT
To me the next logical question is "Why are you trying to solve this equation?" Are you doing it for a class you are taking? If so what types of things have you been talking about in class? What is the section of the book titled? All that info would help to decipher what the objective of the equation is.
Link Posted: 1/30/2006 11:19:43 AM EDT
The answer is 6. I think the problem is trying to confuse you by placing variables and coefficients in unconventional order. If you used the "^" for exponents in other problems to represent exponents, then I would not interpret this equation to have any exponents.
Link Posted: 1/30/2006 11:45:50 AM EDT

Originally Posted By Usafwolfe:
To me the next logical question is "Why are you trying to solve this equation?" Are you doing it for a class you are taking? If so what types of things have you been talking about in class? What is the section of the book titled? All that info would help to decipher what the objective of the equation is.



Yes, it is a question in college for "intermediate algebra." This is week 1. So far, we have discussed silly shit: order of operations, distribution property, terminology, etc. "The value of..." and stuff like that. So I went through the entire text and there is NO example of a question worded like this. Currently, we are discussing chapter 1, which is an overview of algebraic expressions, equations, and principles.

Unfortunately for me, I quit school in 7th grade. I have NO math background except what I learned on my own and verified by getting my GED in '89 to go into the Army. What is a review for some is new to me, and this question makes no sense.

So, the "quiz" for this week contained this question, exactly. And, as usual, I am lost.
Link Posted: 1/30/2006 11:52:33 AM EDT
At face value I would have to say this is an order of operations problem designed to test you on basic logic. If there are no exponents in the problem, as you have it written, then it is a simple matter of inserting (-1) everywhere you see X and then solving the equation according to the proper order of operations.
Link Posted: 1/30/2006 11:53:16 AM EDT
richard

Take a look at the algebraic functions on this web page -
campus.northpark.edu/math/PreCalculus/Algebraic/

This is the conventional method for displaying a value raised to a power. The use of the X^n format came about to accomodate computer programming and it has worked its way into everyday use, such as on pages like this where the word processor does not include a means to show the function as everyone expects it to look in a textbook.

If you are copying this from a book or an instructor's typed notes, there is a fair chance the formatting was lost, hence the equation is not printed as it was originally intended. On the other hand, as I posted above, the way you wrote the function is a great way to test a students understanding.

You need to check with the instructor to determine whether you are looking at a typo. Good luck in your classes.
Link Posted: 1/30/2006 12:01:48 PM EDT

Originally Posted By richardh247:

So, 2x2 COULD mean "two to the power of two times two," "two times X to the power of two," or "two times X times two."



I believe what you are seeing is a typo. An exponent can be expressed as 2x^2 or with a superscripted 2 behind the x. I suspect it is supposed to be superscripted and the formatting was lost somewhere. It is not standard practice to have a term like 2x2 (it would be represented by 4x if it were true). So, my guess is that the correct version is that it is 2 times x-squared or 2 times -1 times -1 = 2.



With the (-1) representing X it doesn't matter. But what is f(-3)?



If you use f(-3) then the answer (assuming the incorrect formatting described above) would be:
-27 + 18 + 15 + 8 = 14

f(x) just means you are performing a function on the variable x. Whatever number is in the parentheses gets substituted for x.

Link Posted: 1/30/2006 12:04:11 PM EDT
Best bet, ask the instructor for the correct format.
Link Posted: 1/30/2006 12:08:28 PM EDT
f(X)= is just another way of writing y=. The f(#) is just telling you what number to substitute for X.

The numbers after the X are understood to be the exponent you are raising the X to. Keep in mind that textbooks, especially math textbooks tend to have mistakes. Half the answers in the back of my calculus textbook are wrong, for example.
Link Posted: 1/30/2006 1:42:30 PM EDT
Thank you, folks! Yes, I mailed the professor, but I haven't heard anything yet so I thought to ask y'all.
Link Posted: 1/30/2006 1:51:53 PM EDT
[Last Edit: 2/1/2006 6:57:14 PM EDT by danpass]

Originally Posted By richardh247:
Thank you, folks! Yes, I mailed the professor, but I haven't heard anything yet so I thought to ask y'all.



He's going to tell you the answer is '87' so get ready .



eta:
It'll turn out that he is an Arfcommer and has been secretly reading this thread .
Link Posted: 2/1/2006 6:03:58 PM EDT

Originally Posted By richardh247:

Originally Posted By Tim84K10:
Don't you have a calculator?

Just press Y= and type in the equation man.



I have a TI BA II Plus, and it has no such button. Besides though, I don't want the calc to do the work. I want to know it - tht's the idea of college: to learn and then pass, not just to pass and walk away without retaining anything. If I can use the calc to check my answers, that would be great! But alas, I don't know what the "Y=" function does, nor does it appear I have one sigh.



You need a graphing calculator for the kind of class you're taking. Being able to see the answers graphically makes this type of math so much easier, trust me. I thought $125 calculator was a bunch of crap until I stated using it. I just got finished with my math homework and it's a piece of cake when you can use the calculator to do some of the work.

Trust me, there is no learning value lost in the use of a calculator. When I was in high school, we never used a graphing calculator. What a mistake that was! In college, it's not only EXPECTED that you know how to use this thing, it makes it so much easier!

I thought I was going to struggle in math forever, but this thing is truely a gift from god. Once you learn how to use it, a graphing calculator is the shit.

Mine is a TI-84 Plus Silver Edition, but there are others that do similar functions. As long as it can graph and table equations at my level of math, it's enough for me.

YMMV.
Link Posted: 2/1/2006 6:14:36 PM EDT
Tag for the reply from the professor.
Link Posted: 2/1/2006 6:25:24 PM EDT
The...answer...to...the...question...is...fourty-two.

Sorry, couldn't resist. Hope the professor squares you away.
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