proof including nonexistent roots

[red and white kop!]

i must prove that this function 2(3x+1) / 3(x^2 - 9) can take on all real values.

f(x)=y=2(3x+1) / 3(x^2 - 9)
=> 3x^2y - 6x - 27y -2=0
=> 36 - 12y(-27y -2) >= 0
=> 27y^2 + 2y + 3 >= 0
=> 27y^2 + 2y + 3 has no real roots
f(x) can take on all real values

is this reasoning conclusive?

introducing y as 'values that can be taken', i group the expression to form a quadratic equation, the roots of which must be larger than or equal to zero for the equation to be real. this inequality in turn forms a quadratic expression with y as a variable, quadratic which is proven to have no real roots itself. from this outcome i find it hard to conclude. is this a proof that f(x) = 2(3x+1) / 3(x^2 - 9) can take on all real values? why? explanations concerning the meaning of nonexistent roots are unclear in my textbook. is this proof wrong or just lacking sufficient links?

 

[Deleted member 4993]

f(x)=y=2(3x+1) / 3(x^2 - 9)

Your problem statement is incorrect. Following your work below - I assume your problem to be:

y = 2 * (3x +1)/[3 * (x^2 - 9)]

And you are trying to prove that

the range of f(x) = ± ?

=> 3x^2y - 6x - 27y -2=0

=> 36 - 12y(-27y -2) >= 0<<< If you are looking at the discriminant of the above quadratic - you should state so. You should not claim that the discriminat is greater that 0 - you are trying to prove it

=> 27y^2 + 2y + 3 >= 0

=> 27y^2 + 2y + 3 has no real roots <<< Correct - however that does not mean that it is always positive - that is what you want to prove.

Since the quadratic above has a positive first term - and it does not have any real root - that specifies that the quadratic is always positive.

f(x) can take on all real values

is this reasoning conclusive? - yes - however as written by you - you would have gotten 50% grade for this problem.

 

[red and white kop!]

please try to follow my written explanation before writing out big red mistakes.
i introduced an inequality to try to find the roots of the quadratic equation, obviously by using the quadratic formula. yes i just realised 12y is wrong. but how can i prove that the function can take on all values? i was given that problem, its not the problem thats in question, its the solution. can i link the absence of real roots to the fact that all values can be taken? i so, how and why? arithmetic blunders are nonimportant here.

 

[stapel]

I'm afraid I don't follow your reasoning. You seem to think that, by proving that the function can never take on the value zero, you have somehow proven that the function can take on any value, which would necessarily include zero. This does not appear to make sense...?

Since you are attempting to do the proof with the algebra and arithmetic, I'm afraid algebraic and arithmetical errors do matter.

To show that a function takes on all y-values, it would be sufficient to invert the function (solve for x in terms of y) and show that this inverse is defined everywhere. Thus, for any y-value, you can find the x-value that gives that y-value, and therefore the function (y) takes on all possible values.

 

[red and white kop!]

so if a quadratic has no real roots, then its sign is determined by that of its first and therefore squared term? so in general this was the only major link missing in my proof? is stapel's method simply an alternative and equally conclusive? wouldnt it be far more complicated to solve for x in this case?

 

[stapel]

I'm sorry, but I have no idea what you're talking about here...?

You have a function: y = [2(3x + 1)] / [3(x^2 - 9)]

You want to show that, for any value y, you can find an x which gives that y-value as the output of the function. So you need to solve for x in terms of y:

3y(x^2 - 9) = 6x + 2

3x^2y - 27y = 6x + 2

3x^2y - 6x - 27y - 2 = 0

(3y)x^2 + (-6)x + (-27y - 2) = 0

Apply the Quadratic Formula, you can find an expression for the required x-value in terms of the desired y-value. Also, you should be able to prove that the discriminant is never negative, so the expression for x is always defined; in addition, you should be able to prove that the discriminant can equal 9, so the square root can equal 3, so the expression can equal zero.

Since all other values follow from the discriminant never being negative, you can thus show that all y-values are possible, using the formula you have created.

 

[Denis]

Hey Sir R.W.Kop, you are close to being as confusing as a teacher ... but not quite :wink:

 

[red and white kop!]

ok thanks everybody, including the werewolf for his very pertinent remark, i will try to improve if i ever become a math teacher