Math....again!

SkyHog

SkyHog

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This one is for rep points :D

Ok - so I think I got the right answer to this, but it seems like it should be more complicated than this....can someone check me to see if I'm wrong?

Given that f(x) = sqrt(x), find f(x)-f(1)/(x-1)

To me, the answer should be 1. Am I right or wrong, and why? This is test review for a test tomorrow.
 
SkyHog said:
This one is for rep points :D

Ok - so I think I got the right answer to this, but it seems like it should be more complicated than this....can someone check me to see if I'm wrong?

Given that f(x) = sqrt(x), find f(x)-f(1)/(x-1)

To me, the answer should be 1. Am I right or wrong, and why? This is test review for a test tomorrow.
Click to expand...

Wow, sorry, I never could think like that. If I can see a physical problem, I can use mathmatics to help to a conclusion. Thing is I could never visualize a problem from lines as above. If I need someone with that kind of ability, I hire them, works out better in the long run.
 
SkyHog said:
This one is for rep points :D

Ok - so I think I got the right answer to this, but it seems like it should be more complicated than this....can someone check me to see if I'm wrong?

Given that f(x) = sqrt(x), find f(x)-f(1)/(x-1)

To me, the answer should be 1. Am I right or wrong, and why? This is test review for a test tomorrow.
Click to expand...
With the equation as written above all I can get is a simple subsitution of

= SQRT(x)-1/(x-1)


But if you forgot a couple of parens so that the original equation looks like this

(f(x)-f(1))/(x-1) then I get a more interesting answer

=(x^1/2 - 1^1/1)/(x-1)

= (x-1)^1/2/(x-1)^1

=(x-1)^(1/2-1)

=(x-1)^-1/2

I have to admit some of my alegbra is a little rusty on this one.
 
smigaldi said:
With the equation as written above all I can get is a simple subsitution of

= SQRT(x)-1/(x-1)


But if you forgot a couple of parens so that the original equation looks like this

(f(x)-f(1))/(x-1) then I get a more interesting answer

=(x^1/2 - 1^1/1)/(x-1)

= (x-1)^1/2/(x-1)^1

=(x-1)^(1/2-1)

=(x-1)^-1/2

I have to admit some of my alegbra is a little rusty on this one.
Click to expand...
Nope - no parenthesis are missing. Looks like i had a big ol' brain fart, and you're right, it doesn't simplify much, does it?

Unfortunately, I cannot give you more reputation. I have given you some somewhere else, and it says I have to spread it around.
 
I couldn't get it simplified any further than scott, and neither could my TI-89...
 
Given that f(x) = sqrt(x), find f(x) - f(1)/(x-1)

sqrt(x) - sqrt(1)/(x-1)

sqrt(x) - 1/(x-1)

sqrt(x) = 1/(x-1)

sqrt(x)*(x-1) = 1

(x^3/2 - x^1/2)^2 = 1^2

x^3 - x - 1 = 0

at this point I must revert to MATLAB, which, upon doing the roots command, gives me an answer of...

1.3247
(And a couple of imaginary answers)


I'm a math minor and I still don't know if that's right. :/
 
wbarnhill said:
Given that f(x) = sqrt(x), find f(x) - f(1)/(x-1)

sqrt(x) - sqrt(1)/(x-1)

sqrt(x) - 1/(x-1)

sqrt(x) = 1/(x-1)

sqrt(x)*(x-1) = 1

(x^3/2 - x^1/2)^2 = 1^2

x^3 - x - 1 = 0

at this point I must revert to MATLAB, which, upon doing the roots command, gives me an answer of...

1.3247
(And a couple of imaginary answers)


I'm a math minor and I still don't know if that's right. :/
Click to expand...

You can't do what I bolded up in the quote. You basically just took 3-4 and made it 3=4 as an example. What was on the other side of the equal sign is unknown NOT zero.
 
Scott,

All he did was add to each side. That's valid algebra.
 
Seems to me that if your asked to "find f(x) - f(1)/(x-1)"

You're looking for a graph of the function not a solution for a particular value. Note that is 2 hyperbolic curves from [0-1) and (1-infinity).

If there's a typo I would expect it to be find f(x) = f(1)/(x-1) which is what William did but I think he made a small mistake.

I get
sqrt(x)(x-1)=1
x (x^2-2x-1) = 1
x^3 - 2x^2 -x -1 = 0

which says x ~= 2.546

Joe
 
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N2212R said:
Scott,

All he did was add to each side. That's valid algebra.
Click to expand...

eh, i dont think so. It wasn't equal to zero to begin with, so there is no other side to add to.

former math nerd
 
N2212R said:
Scott,

All he did was add to each side. That's valid algebra.
Click to expand...
But he did not know what was on the right side of the equal sign. It may not be zero, look at the equation, it is f(x)-f(1)/(x-1) = ?????

William assumed the ??? is zero and the added f(1)/(x-1) to each side. Since we really do not know what is on the right side of the equal sign we have to assume it could be any rational number, lets call it Z.

f(x)-f(1)/(x-1) = Z

So when William decide to add the term f(1)/(x-1) to each side he should have also added it to Z and then got the result of

f(x) = Z+f(1)/(x-1)

NOT

f(x) = f(1)/(x-1)

It is a big difference
 
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I was always under the impression that when you solve for X, you always set the other side =0 At least that's from when I was in school.

Solve for x+3
the answer was always assumed to be -3

If there is no other side, there's nothing to solve for.
But to answer Nick's original question I guess the answer would be all numbers greater than 0 except 1.
 
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One clue is f(x) - sqrt(x) = 0 .

It is left to the reader to solve for x....

N2212R said:
I was always under the impression that when you solve for X, you always set the other side =0 At least that's from when I was in school.

Solve for x+3
the answer was always assumed to be -3

If there is no other side, there's nothing to solve for.
But to answer Nick's original question I guess the answer would be all numbers greater than 0 except 1.
Click to expand...
 
N2212R said:
But to answer Nick's original question I guess the answer would be all numbers greater than 0 except 1.
Click to expand...

greater than or equal to 0 except 1
 
took the test today, and we graded it. The final answer is:

sqrt(x)-1/(x-1) although, one could move the sqrt(x) to the bottom, which was also a correct answer (teacher says that the whole no rationals in the denominator is not valid any more).
 
SkyHog said:
took the test today, and we graded it. The final answer is:

sqrt(x)-1/(x-1) although, one could move the sqrt(x) to the bottom, which was also a correct answer (teacher says that the whole no rationals in the denominator is not valid any more).
Click to expand...

HAHA, so we all just overcomplicated it? Whoops :D
 
I disagree with the zero thing. But, then again, when I was in school, ratios were simply a question of whether we were talking about laden African sparrows, or laden English sparrows. In this case, if there IS no other side, it's not an equation, but a simplified expression--as opposed to a complicated expression. Ah, the good ol' days.


*or was it swallows?
 
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p8cleared2land said:
I disagree with the zero thing. But, then again, when I was in school, ratios were simply a question of whether we were talking about laden African sparrows, or laden English sparrows. In this case, if there IS no other side, it's not an equation, but a simplified expression--as opposed to a complicated expression. Ah, the good ol' days.


*or was it swallows?
Click to expand...

Swallows, african or european.... always nice to find another fan.
 
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