Range of function

**champrock** · June 6th 2009, 02:34 AM

Hi

What will be the range of the function f(x) = 1-1/x .

1. The domain is given as [0, infinity)

2. The domain is given as [0, infinity]

I am primarily confused if "1" will be included or not? The limit of the function tends to 1 but it will never actually reach 1. So, will the range be [-infinity,1) or [-infinity,1] or (-infinity,1) or (-infinity,1]

thanks

**Prove It** · June 6th 2009, 02:58 AM

Originally Posted by champrock

Hi

What will be the range of the function f(x) = 1-1/x .

1. The domain is given as [0, infinity)

2. The domain is given as [0, infinity]

I am primarily confused if "1" will be included or not? The limit of the function tends to 1 but it will never actually reach 1. So, will the range be [-infinity,1) or [-infinity,1] or (-infinity,1) or (-infinity,1]

thanks

Hint: Find the range of the function $f(x) = \frac{1}{x}$ in the domain $x \geq 0$ .

The multiplication by $-1$ gives a reflection in the x-axis, and the addition of 1 is a translation of 1 unit vertically upward.

**Amer** · June 6th 2009, 03:10 AM

I write the answer but when I see the member above me give you a hint I edit it

note you can't write the interval closed from the direction of infinity
like this

(0,infinity]

the correct like this

(0,infinity)

**champrock** · June 6th 2009, 04:20 AM

Originally Posted by Amer

I write the answer but when I see the member above me give you a hint I edit it

note you can't write the interval closed from the direction of infinity
like this

(0,infinity]

the correct like this

(0,infinity)

hmm ok . (0,infinity) is because infinity cant be included in the set right? Its not any proper number.

@Prove It : I found the range. Its coming out to be -infinity to 1. But I am not sure if 1 is actually included or not because
Lim 1/x (x tends to infinity) ---> 0
so 1-1/x = 1-0 = 1
but
But 1/(some very large number ) ----> some very tiny number like 0.000000000045
So, 1-1/x = 1-some very tiny number. = 0.99999999999999434

SO, the range is (-infinity,1) or (-infinity,1] ?

**Amer** · June 6th 2009, 04:22 AM

the first one (-infinity,1)

thanks to prove it

**Prove It** · June 6th 2009, 04:40 AM

Originally Posted by champrock

hmm ok . (0,infinity) is because infinity cant be included in the set right? Its not any proper number.

@Prove It : I found the range. Its coming out to be -infinity to 1. But I am not sure if 1 is actually included or not because
Lim 1/x (x tends to infinity) ---> 0
so 1-1/x = 1-0 = 1
but
But 1/(some very large number ) ----> some very tiny number like 0.000000000045
So, 1-1/x = 1-some very tiny number. = 0.99999999999999434

SO, the range is (-infinity,1) or (-infinity,1] ?

Let's put it this way...

In the domain $x > 0$ , (since $x \neq 0$ as pointed out earlier), the range of the function $\frac{1}{x}$ is $y > 0$ , or in interval notation, $(0, \infty)$ .

Notice that as $x \to \infty, y \to 0$ . BUT since we can never reach $\infty$ , $y$ will never reach $0$ .

Now the reflection in the $x$ axis means that the range of $-\frac{1}{x}$ becomes $y < 0$ , or to use interval notation, $(-\infty, 0)$ .

The translation of 1 unit vertically upward means that the range of $1 - \frac{1}{x}$ is $y < 1$ , or $(-\infty, 1)$ in interval notation.

So to answer your question, you do NOT include the 1 in the range.

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