Domain and Range

1. Jun 21, 2007

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.

2. Jun 21, 2007

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?

3. Jun 21, 2007

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.

4. Jun 22, 2007

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.

5. Jun 22, 2007

radou

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

6. Jun 26, 2007

kuahji

Thanks, this explanation helps.

7. Jun 29, 2007

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.

8. Jun 29, 2007

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.

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