[SOLVED] Domain and Range of this Function

**unstopabl3** · May 7th 2010, 01:25 AM

$f(x)=\frac{6}{2x+3}$ For x is greater than or equal to Zero

How can I find the range of this function? I would normally use the Completing Square Method, but I am not sure how to apply that method in such functions. Can I use it here if so then how?

Help will be appreciated!

Thanks!

**Dinkydoe** · May 7th 2010, 03:09 AM

You mean the range of $f(x)$ for $x> 0$ since $f$ is not defined in $x=0$ .

Have you thought of determining $f'(x)$ ?

If you check carefully you see that the derivative is negative for all $x>0$ wich means that $f$ is a strictly decreasing function. (not very surprising is it?).

And $f(x)> 0$ for all $x$ .

f 's given domain is the open interval $(0,\infty)$ and since as $x\to 0$
we see that $f(x) \to \infty$ .

What can you conclude from this?

**unstopabl3** · May 7th 2010, 04:59 AM

I didn't understand all of that but I've understood most of it.

The correct answer is f>0 and f is less than or equal to 2.

But I am still not sure how to get to this answer.

**Dinkydoe** · May 7th 2010, 05:06 AM

Indeed, the open interval $(3,\infty)$ is the range

What exactly do you mean by f is less then or equal to 2?
It's not, as far as I know. (we just determined that $3<f(x)<\infty$ for all x>0)

Btw, is $f(x)= \frac{6}{2}x+3$ or $\frac{6}{2x}+3$ . It's quite different
I assumed the second case.

(made a typo, hope you see the refreshed version ;p)

Look carefully what happens when $x\to\infty$

**unstopabl3** · May 7th 2010, 05:16 AM

Well here's the question itself.

As the range for a function is the domain of it's inverse, that's why I was trying to figure out the range of the function f to write it down as the domain of it's inverse.

But the correct answer is f>0 and f is less than or equal to 2. ( Domain of f –1: 0 < x ≤ 2 )

Any idea how they got it?

**Dinkydoe** · May 7th 2010, 05:26 AM

haha ok, so actually it's even a different function. It's $f(x)=\frac{6}{2x+3}$ . Be careful next time with your notation, not to leave anything to the imagination of the reader

But yes, now I have a good idea how they got there.

as $x\to\infty$ we see that the denominator becomes very large, while the numerator stays constant 6. This means that $f(x) \to 0$ as x becomes very large. (but never really hits the zero)

While $f(0)= \frac{6}{2\cdot 0 + 3} = 2$

We can not immediately conclude that $0<f(x)\leq 2$ but we're almost there. If we notice that $f'(x)< 0$ for all $x\geq 0$ (check this for yourself) we see that $f$ is strictly decreasing and $\lim_{x\to\infty}f(x)=0$ so $f$ can not take any other values then in the half open interval $(0,2]$ .

**unstopabl3** · May 7th 2010, 05:33 AM

I've got the concept behind it. Thanks!

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