Finding range of function by using algebra?

**dorkymichelle** · December 11th 2009, 06:25 PM

Use algebrad the range of $f(x) = 3x^2+10/x^2-9$
i remember our teacher told us to solve for A, any real number so I did
$3x^2+10/x^2-9 =A$
$3x^2+10=A(x^2-9)$
$3x^2+10=Ax^2-9A$
$9A+10=Ax^2-3x^2$
$9A+10=x^2(A-3)$
$9A+10/A-3 = x^2$
A cannot equal 3
I don't know what the next step is

**earboth** · December 11th 2009, 10:17 PM

Originally Posted by dorkymichelle

Use algebrad the range of $\bold{\color{red}f(x) = (3x^2+10)/(x^2-9)}$
i remember our teacher told us to solve for A, any real number so I did
$3x^2+10/x^2-9 =A$
$3x^2+10=A(x^2-9)$
$3x^2+10=Ax^2-9A$
$9A+10=Ax^2-3x^2$
$9A+10=x^2(A-3)$
$9A+10/A-3 = x^2$
A cannot equal 3
I don't know what the next step is

1. Since $x^2 \geq 0$ for all $x \in \mathbb{R}$ the LHS of the equation must be greater zero too.

2. A quotient is greater (or equal) zero if the signs of denominator and numerator are equal:

$9A+10\geq 0 \wedge A-3>0~\vee~9A+10\leq 0\wedge A-3<0$

$A\geq -\frac{10}9 \wedge A>3~\vee~A\leq -\frac{10}9\wedge A<3$

$\boxed{A>3~\vee~A\leq -\frac{10}9}$

That's the range of the function.

Btw: Please write numerator and denominator of a quotient in brackets. Otherwise you mean: $f(x) = 3x^2+\frac{10}{x^2}-9$

**red_dog** · December 11th 2009, 10:19 PM

$x^2=\frac{9A+10}{A-3}$

The equation must have real solutions. Then the condition is that

$\frac{9A+10}{A-3}\geq 0\Rightarrow A\in\left(\left.\-\infty,-\frac{10}{9}\right]\right.\cup(3,\infty)$

**dorkymichelle** · December 11th 2009, 10:57 PM

I understand the part where it has to be greater than 3, because that would make the demoniator zero. But i don't understand why there is a restriction on -10/9

**11rdc11** · December 11th 2009, 11:34 PM

Originally Posted by dorkymichelle

I understand the part where it has to be greater than 3, because that would make the demoniator zero. But i don't understand why there is a restriction on -10/9

Think of it this way. You found your critical points to be 3 and -10/9 by setting the numerator and denominator to 0

Now test a value from - infinity to -10/9, next a value from -10/9 to 3, and then from 3 to positive infinity.

You will notice that end up with a positive value in the intervals of (- infinity to -10/9) and (3 to positive infinity).

You will end up with a negative value in the interval of (-10/9 to 3)

Now since you can only have real solutions you can't use the negative interval.

The reason why you can't use the negative interval is when you take a square root of a negative number you get an imagnery number.

Does this make sense?

**dorkymichelle** · December 11th 2009, 11:58 PM

So basicly, 9A+10/A-3 cannot be a negative number because you can't square root a negative number and still have a real number, because of $x^2$ and -10/9 is the number that makes the fraction 0 and any number less than that would make the fraction negative correct?

**11rdc11** · December 12th 2009, 12:21 AM

Originally Posted by dorkymichelle

So basicly, 9A+10/A-3 cannot be a negative number because you can't square root a negative number and still have a real number, because of $x^2$ and -10/9 is the number that makes the fraction 0 and any number less than that would make the fraction negative correct?

Close but not quite.

Yes

$\frac{9A+10}{A-3}$

has to be positive but the solution set is like red dog said. Notice how that includes values less than $\frac{-10}{9}$

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