Pre-Calculus help !!
- Thread starter TimSon42
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I'm sorry, but "1 / (x + 1) / 2" has no standard mathematical meaning. Did you intend one of the following?
. . . . .\(\displaystyle \dfrac{1}{\left(\dfrac{x\, +\, 1}{2}\right)}\)
. . . . .\(\displaystyle \dfrac{\left(\dfrac{1}{x\, +\, 1}\right)}{2}\)
Also, you state that you are supposed to "evaluate" the expression, but this can only be done after one has been provided a value for the variable. What value did they give you?
When you reply, please include a clear listing of your efforts so far. Thank you!
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To "evaluate" an expression, it is necessary to have values for any variables. What values did they give you?
Please confirm or correct:
. . . . .\(\displaystyle 2\, +\, \dfrac{\left(\dfrac{1}{x\, +\, 1}\right)}{2}\, -\, \dfrac{1}{x\, -\, 1}\)
To "solve", one must have an equation (something with an "equals" sign in the middle of it). You did not include this information in your post. Please provide the entire equation (that is, the posted expression, the "equals" sign, and whatever is on the other side of the "equals" sign) and the correct instructions.
How did you get this? Note: It cannot be a valid simplification ("solution"?), since the two expressions do not have the same domain. For instance, x = 0 is allowed in the original expression, but not in your "solved" "equation".
Please be complete. Thank you.
To "evaluate" an expression, it is necessary to have values for any variables. What values did they give you?
Please confirm or correct:
. . . . .\(\displaystyle 2\, +\, \dfrac{\left(\dfrac{1}{x\, +\, 1}\right)}{2}\, -\, \dfrac{1}{x\, -\, 1}\)
To "solve", one must have an equation (something with an "equals" sign in the middle of it). You did not include this information in your post. Please provide the entire equation (that is, the posted expression, the "equals" sign, and whatever is on the other side of the "equals" sign) and the correct instructions.
How did you get this? Note: It cannot be a valid simplification ("solution"?), since the two expressions do not have the same domain. For instance, x = 0 is allowed in the original expression, but not in your "solved" "equation".
Please be complete. Thank you.
There are no given X values...
On Wolfram, it shows that the correct root answer is X = -3/2
D
Deleted member 4993
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Okay; so then the instructions could not have been to "evaluate". What were the actual instructions?
Please reply with the typed-out text of whatever was supposed to be in the image.(there's supposed to be an image here: ==><== ...but it doesn't display, and it possibly doesn't exist)
What do you mean by a "correct root answer"? Were you supposed to be working with the following instead of what you'd posted?
. . . . .\(\displaystyle f(x)\, =\, 2\, +\, \dfrac{\left(\dfrac{1}{x\, +\, 1}\right)}{2}\, -\, \dfrac{1}{x\, -\, 1}\)
Were the instructions to "find the roots of the function"? What did you enter into Wolfram Alpha in order to obtain your result?
Please keep in mind that we cannot see whatever it is that you're looking at. You have to tell us what the actual question is, what the actual instructions are, and what actual steps you actually took. Only then will we have enough information to be able to proceed.
Thank you.
Here is what I typed in Wolfram. (2+(1/(x+1)))/ (2-1/(x-1))
Here are the steps I took.
Worked the numerator:
1) (2(x+1)) / (x+1) + (1/(x+1))
2) (2x+2) / (x+1) + 1/(x+1)
Result:
3) (2x+3) / (x+1)
Then worked the denominator:
1) (2(x-1)) / (x-1) - (1/(x-1))
2) (2x-2) / (x-1) - (1/(x-1))
Result:
3) (2x-3) / (x-1)
Then I multiplied the reciprocal of both results:
1) (2x+3) / (x+1) * (x-1) / (2x-3)
2) (2x+3) / (2x-3)
And this is where I couldn't find the root answer. "X = -3/2"
D
Deleted member 4993
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You have not answered the most important question asked by Stapel:
What was the correct instruction (question)?
Was it:
Solve for the root/s of the given function?
Please answer this first!!
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And what did you ask Wolfram's Alpha to do?
Also, this is the third or fourth version of the expression. What, exactly, is the actual expression?
"Evaluating" means "plugging the given number(s) into the specified variable(s), and simplifying to find the numerical result." If the instructions were to "evaluate", then you cannot be expected to find a value for x (that is, find roots, solve an equation, etc); you are expected to evaluate for a given value of x. What value did they give you?
Bob Brown MSEE
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y=0 when x=-3/2
The problem should be stated as ...
Find the zero(s) of y, where
y = (2 + (1/(x + 1)))/(2 - (1/(x - 1)))
As you can see,
Click Here
This function seems to have two zeros at
x = -3/2 and x = 1
However, there is a "hole" in the function at x = 1 because we divide by x-1 during the evaluation of y.
It is possible to "heal" the hole.
Click Here
Using (x-1)(2x+3)/(x+1)/(2x-3)
The problem should be stated as ...
Find the zero(s) of y, where
y = (2 + (1/(x + 1)))/(2 - (1/(x - 1)))
As you can see,
Click Here
This function seems to have two zeros at
x = -3/2 and x = 1
However, there is a "hole" in the function at x = 1 because we divide by x-1 during the evaluation of y.
It is possible to "heal" the hole.
Click Here
Using (x-1)(2x+3)/(x+1)/(2x-3)