Domain & Range of 2x^2 + 4x - 3

[kuahji]

Find the domain & range of the function.

2x^2 + 4x - 3

I attempted to solved doing the following

2x^2 + 4x = 3
x^2 + 2x = 3/2 (divided by two)
(x+1)^2 = 5/2 (completed the square & added one to both sides)
(x+1)^2 - 5/2 = 0

So I put the range was (-5/2, infinity), but the book has it (-5, infinity). It seems any problem where the leading coefficient is greater than one, I'm getting incorrect answers. So there must be an error in how I'm trying to solve the problem.

 

[neutrino]

I don't understand why you have equated the function to zero. As it stands, the range and domain is the real line, assuming x and f(x) are real. Are there any conditions on x and/or f?

 

[kuahji]

Figured out the way to do it, sorry.

2x^2 + 4x -3
2x^2 + 4x = 3
2(x^2 + 2x) = 3
2(x+1)^2 = 5
2(x+1)^2 - 5

Even though now I'm kinda curious as to why the first method I tried was incorrect. Like why can't you just divide the whole thing by two.

 

[Gib Z]

Well the first time you tried you Assumed that it was equal to zero. And because of that, you changed it to another polynomial, where the co efficients are divided by 2. It has the same ZERO's but different values for other things. The second time you tried you didn't assume anything, you just wrote the expression in another completely equivalent way.

 

[radou]

As Gib Z suggested, "2x^2 + 4x - 3" means nothing. Write "f(x) = 2x^2 + 4x - 3" to avoid primary confusion.

 

[kuahji]

Thanks, this explanation helps.

 

[i.mehrzad]

Well i think it makes sense equating the derivative to zero so that you can find out wether there is a maxima or minima.

After finding the value of x for which there is a maxima or minima then you will have to find the value of the maxima or minima.

 

[morson]

Derivatives should not even be considered when determining the domain and range of a real-valued quadratic function. There exists quite an easy way to express f(x) = ax^2 + bx + c, a \not=\ 0 in the form f(x) = a(x - h)^2 + k. If a is negative, then k is the maximum value of the function, and if it is positive, then k represents the minimum value of the function. The "completing the square" method really is much quicker.