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Original post by Zipster
The understood range of the square root operation is all non-negative numbers. Technically, the "square root" of 4 is 2 and -2, but when you do the operation, it''s assumed to be the positive root only (2). The +/- is simply there to indicate that the negative square root should also be used, since it is mathamatically true, yet simply not assumed when you do the operation.

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Well, sometimes, although mathematically true, the +- square root should be considered, because some equations have two legitimate answers. The - part of the square root should not be ignored. Imagine you have a problem in which someone is 10 meters below ground, and he throws a ball up and wants to see when it reaches the height 0 meters(since he is 10 meters below ground). Technically, using the equation d = vt + .5at^2 will result in two possible solutions, so the + and - of a square root must be considered.

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Well, sometimes, although mathematically true, the +- square root should be considered, because some equations have two legitimate answers. The - part of the square root should not be ignored.

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Nowhere did I say it should be ignored, I merely said that the square root operation yields positive roots. Hence the +/- symbol to indicate the need to consider the fact that there can be negative roots as well (not to say that +/- is used exclusively when dealing with roots). Most of the time the +/- symbol is excluded, however that doesn''t mean the square root operation behaves any differently. In the case of the quadratic equation, the +/- is explicitly used not only because it represents an actual operation, but also to compensate for the fact that there are negative roots

Happy new year everyone.

The reason quadratics don't have a 'proper' inverse function is because a mathematical function has only one result at a given evaluation point. You can say sqrt is the inverse function of x*x if you restrict the range of the inverse to positive square roots only. Otherwise, there are 2 values and hence it aint a function.

sqrt(x) = inverse (x*x) with domain R and range [ 0, +inf )
-sqrt(x) = inverse (x*x) with domain R and range ( -inf, 0 ]

An inverse of a function which can be graphed 2d is the reflection of the graph in the line y = x. If the resulting reflection has only one value for each evaluation point in the domain, it's a function. If it hasn't, you need to make a restriction in the range.

Hope that helps.

EDIT: This post and the post it replies to have absolutely nothing to do with second derivatives though do they?

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

[edited by - Paradigm Shifter on January 2, 2003 5:49:54 AM]

It actually doesn''t but the information is great... I love input!

Now I''m hungry...

Thanx for the replies fellaz...

-Lewis [m80]
Play QUADz MX @
www.m80produxions.com

To find the inverse of a quadratic function, it''s easier to put it in its canonical form (I''m not sure that''s how it''s called in English):

y = b(x - h)² + k

where the maximum/minimum of the parabola is at position (h, k)

To find h, you can find the two zeros of the function; h is the average of the two. k = f(h), and b can be found with any other value.

Normally, though, we use square completion to find the canonical equation.

Once you have the canonical equation, isolating x is trivial.

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Original post by Zipster
The understood range of the square root operation is all non-negative numbers. Technically, the "square root" of 4 is 2 and -2, but when you do the operation, it''s assumed to be the positive root only (2). The +/- is simply there to indicate that the negative square root should also be used, since it is mathamatically true, yet simply not assumed when you do the operation.

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Are you sure about that? From what I''ve learned, the square root operation is a function; when we draw its graph, we only draw the upper branch; not both.

x² = 4 => x = 2 or x = -2

sqrt(4) = x => x = 2

That''s what I learned. If you want both the positive and negative roots, you need to put the +- signs explicitly... Anyway, it''s just semantic...

Cédric

I just made some of my math homework(Yeah, sorry this is a homework question but it''s going about this topic..) and it goes about first derivatives..

Say you have something like 3x^4+x^3+x^2 and you need to give the extreme values of such a function then they thought me to do:

F''(x)=12x^3+3x^2+2x
And then solve: F''(x)=0.. But how to do that? The way I can solve a quadratic equation with that x = (-b +- sqrt(D))/2a that only applies to ^2 functions and not to ^3..

In my math book there are several cases with ^3 equations but I have no clue how to solve them.. Something I need to know or is it just not solvable with my current knowledge?

I hope this question is allowed.. if not I''m sorry

you can try factoring out all the x''s from the polynomial and then setting each term to zero... if not, keep factoring.

I think it''s time this thread is closed either way... thanx for the help fellaz.

-Lewis [m80]
Play QUADz MX @
www.m80produxions.com

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Are you sure about that?

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Yup.

I probably shouldn''t have used the word "assumed", and I should have used the word "function" instead of "operation", but basically I was saying exactly what you said.

Scheermesje: I''m sure that if you tried you''d find a common factor of F''(x)=12x^3+3x^2+2x

Then you''ll have two parts to F''(x), lets call them G(x) and H(x)

If you''re solving for G*H = 0 then solutions are G(x) = 0 and H(x) = 0. Solve them seperately for x and you''re finished.

You should end up with 3 possible values for x as F''(x) is a cubic equation. You can look at G(x) as being a scaling factor for H (and of course vice versa) which varies with x.

These kind of multiplications of functions become interesting when modelling physical effects such as damping of oscillations. You may have an function that represents a forwards and backwards motion which is regular and goes on forever. Then you can multiply that function by another which is a straight line decreasing. This would give you the motion of, for example, a yo-yo being used by someone on a descending escalator. Or multiplying instead by a function which is a line decreasing eventually to zero at infinity. This will modify the oscillation to produce a gradual decay in amplitude (as it''s heading toward multiplying the original function by zero). This is like a childrens swing which gradually will stop swinging (if you''re not kicking your legs)

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Original post by petewood
Scheermesje: I''m sure that if you tried you''d find a common factor of F''(x)=12x^3+3x^2+2x

Then you''ll have two parts to F''(x), lets call them G(x) and H(x)

If you''re solving for G*H = 0 then solutions are G(x) = 0 and H(x) = 0. Solve them seperately for x and you''re finished.

You should end up with 3 possible values for x as F''(x) is a cubic equation. You can look at G(x) as being a scaling factor for H (and of course vice versa) which varies with x.

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Thnx I tried it already but I made a silly mistake.. after asking it to a classmate of mine we found at what the mistake was and I solved it

Thnx for your examples to.. We''re now getting things like k(x)=f(x)*g(x) and then you need to calculate k''(x) or with a division but I can''t see the use of it yet..