Range of an equation

minisue1

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Hi, I have this question, "Find the range of (2x^2)/(x^2+x+1)."

What is the best way to go about this? Should I graph it, or can I work it out algebraically?

HELP PLEASE...

whoops, the numerator should be (2x^2+2)
 
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Hi, I have this question, "Find the range of (2x^2)/(x^2+x+1)."

What is the best way to go about this? Should I graph it, or can I work it out algebraically?

HELP PLEASE...
Vocabulary point. This is a rational function, not an equation.

To graph it, you would first have to work it out algebraically. Graphing rational functions is not so simple as graphing lines.

Start by finding the y intercept algebraically.

Finding the y intercept is easy. What does the function equal when x = 0?

OK Now comes a slightly hard part. 2x^2 is obviously never negative.


So \(\displaystyle \dfrac{2x^2}{x^2 + x + 1}\) is never negative if \(\displaystyle x^2 + x + 1 > 0\ for\ all\ x.\)

Can you show that? How?

Now comes the really hard part. Does the function have a finite maximum?

Denis suggests the clever idea of saying it does and solving for it using the quadratic formula. He calls the maximum k.

\(\displaystyle \dfrac{2x^2}{x^2 + x + 1} = k \implies 2x^2 = kx^2 + kx + k \implies\)

\(\displaystyle (k - 2)x^2 + kx + k = 0 \implies x = \dfrac{- k \pm \sqrt{k^2 - 4(k - 2)k}}{2(k - 2)} = \dfrac{- k \pm \sqrt{k^2 - 4k^2 + 8k}}{2k - 4} = \dfrac{- k \pm \sqrt{8k - 3k^2}}{2k - 4}.\)

For x to be a real number, what can we say about k? Well obviously it must be a positive number. What else can we say about it?

That is a KILLER problem for first year algebra.
 
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Low is obviously 0.


Let (2x^2)/(x^2+x+1) = k
That'll lead you to this quadratic:
x = [-k +- SQRT(2k - k^2)] / [2(k - 4)]

SQRT(2k - k^2) : max k = ?
Oh Denis boy

The corner's calling you.

\(\displaystyle \dfrac{2x^2}{x^2 + x + 1} = k \implies (k - 2)x^2 + kx + k = 0.\)

So the discriminant is \(\displaystyle k^2 - 4(k - 2)k = k^2 - 4(k^2 - 2k) = k^2 - 4k^2 + 8k = 8k - 3k^2 \ne 2k - k^2.\)
 
\(\displaystyle (k - 2)x^2 + kx + k = 0 \implies x = \dfrac{- k \pm \sqrt{k^2 - 4(k - 2)k}}{2(k - 2)} = \dfrac{- k \pm \sqrt{k^2 - 4k^2 + 8k}}{4k - 8} = \dfrac{- k \pm \sqrt{8k - 3k^2}}{4k - 8}.\)

For the last two fractions, the denominators should be "2k - 4." \(\displaystyle \ \ \ \ \ 2(k - 2) \ \ne\ 4k - 8.\)
 
If \(\displaystyle \dfrac{2x^2+2}{x^2 + x + 1} = f(x)\)
then f(x)-4 has a double root in the numerator at x=-1

So, I believe the max is f(-1)=4 which would be the upper bound on the range.
 
If \(\displaystyle \dfrac{2x^2+2}{x^2 + x + 1} = f(x)\)
then f(x)-4/3 has a double root in the numerator at x=1

So, I believe the min is f(1)=4/3 which would be the lower bound of the range.
 
Bob

The function had a numerator of 2x^2, not 2x^2 + 2. Denis was correct that the minimum is zero.

This seems to be a problem that most of us insist on doing wrong: I for example forgot how to multiply by 2.

Actually, the OP edited his original post this morning to say the numerator is actually 2x² + 2.
 
So when i end up with the discriminant, and then factorise that,how do i determine whether the range is between, or outside of the roots
 
So when i end up with the discriminant, and then factorise that,how do i determine whether the range is between, or outside of the roots
\(\displaystyle f(x) = \dfrac{2x^2 + 2}{x^2 + x + 1}.\)

Let's go back to denis's clever idea.

\(\displaystyle Let\ f(x) = \dfrac{2x^2 + 2}{x^2 + x + 1} = k \ in \mathbb\ R \implies kx^2 + kx + k = 2x^2 + 2 \implies (k - 2)x^2 + kx + (k - 2) = 0.\)

Make sense so far?

Now using the quadratic formula

\(\displaystyle x = \dfrac{- k \pm \sqrt{k^2 - 4(k - 2)(k - 2)}}{2(k - 2)} = \dfrac{- k \pm \sqrt{k^2 - 4(k^2 - 4k + 4)}}{2k - 4} = \dfrac{- k \pm \sqrt{k^2 - 4k^2 + 16k - 16}}{2k - 4} = \dfrac{- k \pm \sqrt{-3k^2 + 16k - 16}}{2k - 4}.\)

The only way that x can be a real number is if \(\displaystyle - 3k^2 + 16k - 16 \ge 0.\)

Use the quadratic formula again.

\(\displaystyle -3k^2 + 16k - 16 = 0 \implies k = \dfrac{-16 \pm \sqrt{(-16)^2 - 4(-3)(-16)}}{2 * (-3)} = \dfrac{-16 \pm \sqrt{256 - 192}}{-6} = \dfrac{16 \mp \sqrt{64}}{6} = \dfrac{16 \mp 8}{6} \implies\)

\(\displaystyle k = \dfrac{16 - 8}{6} = \dfrac{8}{6} = \dfrac{4}{3},\ or\ k = \dfrac{16 + 8}{6} = \dfrac{24}{6} = 4.\)

So, as Bob Brown said, the range is \(\displaystyle \left[\dfrac{4}{3},\ 4\right].\)
 
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