# [SOLVED] Simple Functions

• September 25th 2009, 08:59 AM
unstopabl3
[SOLVED] Simple Functions
I've got a few questions regarding Functions which are bothering me.

Q1: Find the largest possible domain of each of the following functions:

a) $1/x-2$

Answer = All real numbers except 2

Why is this? Also why can't we use negative integers for x e.g -2

b) $1/(x-1)(x-2)$

Answer = All real numbers except 1 and 2

Why is this? Also why can't we use negative integers for x e.g -2 (i guess it's same as the part (a) I posted)

Q2: The domains of these functions are the set of all positive real numbers. Find their ranges:

a) $f(x) = (x-1)^2 + 2$

Generally I'd put the least real number value (0) into the function and get the least range. In this case my answer is not matching with the real answer.

Answer = f(x) is greater than or equal to 2

Q3: Find the range of each if the following functions. All the functions are defined for all real values of x:

a)
$f(x) = 3(x+5)^2+2$

f(x) is greater than or equal to 2How is that?

b)
$f(x) = 2(x+2)^4-1$

Answer = f(x) is greater than or equal to -1

How is that?

• September 25th 2009, 09:27 AM
Lujan
Q1: If $x=-2$, you've got $1/0$, which is undefined. Any other value of x gives you a defined output. The logic is the same for (b); if either term in the denominator is 0, the product is 0, so you get $1/0$ again.

Q2: $(x-1)^2$ must be 0 or larger, since you're squaring the value at the end (a negative value inside the parenthesis would become positive after squaring it).

The least that term can be is therefore 0, when $x=1$. In that case $f(1)=(1-1)^2+2=2$. Any other value for x gives you a value larger than 2 (try plugging in a few values if you don't see that).

Q3: This is same logic as Q2. The term which is squared can only ever be 0 (if $x=-5$ for (a) or $x=-1$ for (b)) or larger. If you set it to 0, the function evaluates to 2 (for a) or -1 (for b).
• September 26th 2009, 03:56 AM
unstopabl3

For Q1

a) If i put x = -2 doesn't that make the whole fraction

1/-2-2 which is 1/-4 No? Obviously this is not equal to 1/0 so it is a defined value. Real numbers are all positive numbers including zero right?

Please elaborate here as how you got 1/0 by putting x = -2

b) Okay, I know that why we can't put 1 or 2 in here as it will make the fraction 1/0 which is undefined. But again my question why can't we use negative values for x here e.g If i put x = -2 doesn't that make the whole fraction

1/(-2-1)(-2-2) which is 1/(-3)(-4) = 1/12 isn't this a defined value?? Then why can't we use the negative values of x here?

Q2

I understand the working of this type of question but I don't understand why do we have to start the input value of x at 1 and why we can't use x = 0 in this equation? Isn't zero a positive real number?

If I put x = 0 then

f(0) = (0-1)^2 + 2 = 3 so my answer is f(x) is greater than or equal to 3

What am I doing wrong? When you say "The least that term can be is therefore 0", why?

Similarly I don't understand what you are trying to say in Q3.

Please elaborate how you are supposed to tackle these kind of questions. Maybe the method I'm doing is wrong.
• September 26th 2009, 04:22 AM
mr fantastic
Quote:

Originally Posted by unstopabl3

For Q1

a) If i put x = -2 doesn't that make the whole fraction

1/-2-2 which is 1/-4 No? Obviously this is not equal to 1/0 so it is a defined value. Real numbers are all positive numbers including zero right?

Please elaborate here as how you got 1/0 by putting x = -2 Mr F says: It should be obvious that Lujan made a typo. S/he meant x = 2.

b) Okay, I know that why we can't put 1 or 2 in here as it will make the fraction 1/0 which is undefined. But again my question why can't we use negative values for x here e.g If i put x = -2 doesn't that make the whole fraction

1/(-2-1)(-2-2) which is 1/(-3)(-4) = 1/12 isn't this a defined value?? Then why can't we use the negative values of x here? Mr F says: You can. Who said you can't? x = 1 and x = 2 are the only values that are not allowed.

Q2

I understand the working of this type of question but I don't understand why do we have to start the input value of x at 1 and why we can't use x = 0 in this equation? Isn't zero a positive real number? Mr F says: No it's not. However, you can consider the limit of the function as x --> 0.

If I put x = 0 then

f(0) = (0-1)^2 + 2 = 3 so my answer is f(x) is greater than or equal to 3

What am I doing wrong? When you say "The least that term can be is therefore 0", why? Mr F says: Is x = 1 an allowed value? What is f(1)? You should be familiar with the graph of a parabola. Draw a graph of y = f(x) over the stated domain of all positive real numbers. What do you see?

Similarly I don't understand what you are trying to say in Q3. Mr F says: Draw the graphs.

Please elaborate how you are supposed to tackle these kind of questions. Maybe the method I'm doing is wrong.

The best way of finding the range of a function is to draw its graph.
• September 26th 2009, 09:16 AM
Lujan
Whoops, sorry for the typo in Q1.
• September 26th 2009, 10:02 AM
Defunkt
@unstopabl3: You're confusing the set of natural numbers with the set of real numbers.

To clarify:
Natural number - Wikipedia, the free encyclopedia
Real number - Wikipedia, the free encyclopedia
• September 26th 2009, 11:00 AM
unstopabl3
Thanks, much appreciate the help guys. I will work on the above solutions and get back to you.

Keep up the good work ;)

======================

Edit

Mr Fan.

You said that 0 is not a positive real number real but it is a real number right? Just not positive?
Also what do you mean by "However, you can consider the limit of the function as x --> 0"

Lastly you said that one should draw the graphs of the functions to find their ranges, but we can't do that for each function as it would take too much time and would not be accurate. Please correct me if I am wrong.

Thanks again.
• September 26th 2009, 03:05 PM
mr fantastic
Quote:

Originally Posted by unstopabl3
Thanks, much appreciate the help guys. I will work on the above solutions and get back to you.

Keep up the good work ;)

======================

Edit

Mr Fan.

You said that 0 is not a positive real number real but it is a real number right? Mr F says: Yes. Just not positive? Mr F says: Correct.

Also what do you mean by "However, you can consider the limit of the function as x --> 0" Mr F says: x = 0 is not in the domain but x can approach 0 as a limit. Don't worry about this as it will probably be more confusing than helpful. When drawing the graph, just calculate f(0) and then put an open circle at the point to indicate that it's not included.

Lastly you said that one should draw the graphs of the functions to find their ranges, but we can't do that for each function as it would take too much time and would not be accurate. Please correct me if I am wrong. Mr F says: It shouldn't take very long to draw a parabola that is written in turning point form. You only want to use the graph to get the range, so you don't have to worry about things like x-intercepts. ONly show that informtaion that is relevant to finding the range. As I said, you should be able to do this in short order (if not, you need to start practicing).

Thanks again.

Drawing a graph is the best way of getting the range of a function.
• September 27th 2009, 06:37 AM
unstopabl3
Thanks for the quick replies sir.

If 0 is not a positive nor a negative number then how come it's stated on wikipedia as following:

"according to the traditional definition or the set of non-negative integers {0, 1, 2, ...}"

Natural_Numbers

This means that according to the traditional definition 0 is a positive integer. I am pretty sure I heard my A Level Maths teacher say that 0 is a positive integer.

Sir that's exactly what I lack, I need practice on sketching quadratic.

Could you please point me to an easy to follow guide or give me a few tips on how to sketch graphs especially of the following types:

$
y = x^3/2$

$y = x^-4/3$

$y = x^2/3 - x^-2/3$

$y = x^2 + x -4$

$y = x^2(x-4)$

$y = -(x+6)(x+4)(x+2)$

I am familiar with the rules of parabolas (the graphs of y = ax^2+bx+c and what these variables mean). I just get confused when I see factorized forms. Please tell me how to make these graphs without taking a lot of time. I am willing to absorb and practice what ever you throw at me ;)

• September 27th 2009, 08:19 AM
e^(i*pi)
Quote:

Originally Posted by unstopabl3

$y = x^2 + x -4$

$y = x^2(x-4)$

$y = -(x+6)(x+4)(x+2)$

I am familiar with the rules of parabolas (the graphs of y = ax^2+bx+c and what these variables mean). I just get confused when I see factorized forms. Please tell me how to make these graphs without taking a lot of time. I am willing to absorb and practice what ever you throw at me ;)

You went from a quadratic in step one to a cubic in step two...

To factorise think which numbers add to 1 and multiply to -4. Although don't think for too long because there are none. Use the quadratic formula to find x:

$
x = \frac{-1\pm \sqrt{1^2-(4)(1)(-4)}}{2(1)} = \frac{-1 \pm \sqrt{17}}{2}$

To plot the graph find a rough decimal approximation and plot it as $(x_1,0)$, $(x_2,0)$ and $(0,f(0)) = (0,c)$. You can either find more pairs or join up using the correct shape.

Better yet, use a plotting program or a spreadsheet program to get a graph done (see attachment for $x^2+x-4=0$
• September 27th 2009, 09:06 AM
Defunkt
Quote:

Originally Posted by unstopabl3
Thanks for the quick replies sir.

If 0 is not a positive nor a negative number then how come it's stated on wikipedia as following:

"according to the traditional definition or the set of non-negative integers {0, 1, 2, ...}"

Natural_Numbers

This means that according to the traditional definition 0 is a positive integer. I am pretty sure I heard my A Level Maths teacher say that 0 is a positive integer.

You should note that non-negative is not necessarily positive. Actually, 0 is neither negative nor positive. Thats why when we say non-negative integers we mean $\left\{0,1,2,...\right\}$, however the positive integers are $\left\{1,2,3,...\right\}$
• September 28th 2009, 11:19 AM
unstopabl3
Thanks fellas, I will try out the solutions and get back to you ;)