- #1

F(x)=square root of x^2-49

Thanks for your help.

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- #1

F(x)=square root of x^2-49

Thanks for your help.

- #2

Lonewolf

- 337

- 1

- #3

like what number could I plug in there?

- #4

Lonewolf

- 337

- 1

- #5

- #6

Lonewolf

- 337

- 1

Since f is a function from the reals to the reals (assumedly), we need (x

- #7

- #8

Lonewolf

- 337

- 1

To find the domain, try to find where the function "breaks down". As practise, find the domain of f(x) = 1/(x-1) for x a real number. Think about what value of x would make f(x) make no sense.

- #9

I have no idea. I'm totally lost. This is due in less than an hour. Please help me out. Thanks

- #10

Lonewolf

- 337

- 1

The solution is the domain of x is |x| > 7.

Why? Well, we have f(7) = f(-7) = 0. If we set x = 7, and then decrease x slightly, f(x) turns into an imaginary number, and since we're working with reals, this is of no use to us. Similarly for x = -7, if we increase x slightly, we again get an imaginary number. Since (x

With respect to your teacher, isn't there another one you could ask? It sometimes helps a lot to get more than one explanation for something, particularly if you can't understand the first...

- #11

agro

- 46

- 0

Btw, Lonewolf meant |x| [>=] 7 (the absolute value of x must be equal to or greater than 7), since at x = 7 or x = -7, the function value is [squ]0 = 0.

Another way of saying "|x| [>=] 7" is "x [<=] -7 or 7 [>=] 7"

Btw, I'll try to give an explanation of a function's natural domain. If it is unclear or wrong please point me out.

--- natural domain of a function ---

[note that we are discussing a function that maps a real number to another real number]

The domain of a function can be stated explicitly, such as

f(x) = 1, x >0

The domain of the function above is all real numbers greater than 0. With that restriction, f(0), f(-2), f(-3.43), etc are undefined.

If a function has no domain stated explicitly, then it is understood that the domain consists of all real number x for which f(x) is defined. This implicit domain is called the natural domain of f(x).

Let's look at the function f(x) = 1/x

for x = 1, f(x) = 1/1 = 1 (defined). For x = 5, f(x) = 1/5 (defined also). It turns out that only at x = 0 is f(x) undefined. That is because 1/0 is undefined. So the domain of the function f(x) = 1/x is all x values other than 0.

How about the function f(x)=[squ]x?

At x equal or greater than 0, f(x) is defined. At x less than 0, f(x) is undefined. For example, [squ](-1) is undefined (because there are no real number y that satisfies the property y^2 = -1). That means the natural domain of the function f(x) = [squ]x is all x equal to or greater than 0.

Now, let's go back to your problem. Your function is f(x)=[squ](x^2-49). Let's introduce a new variable b, where b = x^2-49. We can then rewrite your function as f(x) = [squ]b. We already know from the previous explanation that in order to make f(x) defined, b must be equal to or greater than 0. So we write

b [>=] 0

But b = x^2-49, so we can write

x^2-49 [>=] 0

By factoring, we get

(x+7)(x-7) [>=] 0

At x = 7 or x = -7, the value of b = (x+7)(x-7) will be zero (do you know why?). Since [squ]0 is defined, f(x) is defined at these x = 7 and x = -7. At values of x between 7 and -7, the value of b will be less than 0 (verify this). That means f(x) is undefined for -7 < x < 7. For x < -7 or x > 7, b is positive (verify this also) and that means f(x) is defined.

Thus we can conclude that the natural domain of f(x) consists of all x such that

x is equal to -7, or

x is smaller than -7, or

x is equal to 7, or

x is greater than 7

We can write this as |x| [>=] 7.

Does every part of my explanation makes sense to you?

Another way of saying "|x| [>=] 7" is "x [<=] -7 or 7 [>=] 7"

Btw, I'll try to give an explanation of a function's natural domain. If it is unclear or wrong please point me out.

--- natural domain of a function ---

[note that we are discussing a function that maps a real number to another real number]

The domain of a function can be stated explicitly, such as

f(x) = 1, x >0

The domain of the function above is all real numbers greater than 0. With that restriction, f(0), f(-2), f(-3.43), etc are undefined.

If a function has no domain stated explicitly, then it is understood that the domain consists of all real number x for which f(x) is defined. This implicit domain is called the natural domain of f(x).

Let's look at the function f(x) = 1/x

for x = 1, f(x) = 1/1 = 1 (defined). For x = 5, f(x) = 1/5 (defined also). It turns out that only at x = 0 is f(x) undefined. That is because 1/0 is undefined. So the domain of the function f(x) = 1/x is all x values other than 0.

How about the function f(x)=[squ]x?

At x equal or greater than 0, f(x) is defined. At x less than 0, f(x) is undefined. For example, [squ](-1) is undefined (because there are no real number y that satisfies the property y^2 = -1). That means the natural domain of the function f(x) = [squ]x is all x equal to or greater than 0.

Now, let's go back to your problem. Your function is f(x)=[squ](x^2-49). Let's introduce a new variable b, where b = x^2-49. We can then rewrite your function as f(x) = [squ]b. We already know from the previous explanation that in order to make f(x) defined, b must be equal to or greater than 0. So we write

b [>=] 0

But b = x^2-49, so we can write

x^2-49 [>=] 0

By factoring, we get

(x+7)(x-7) [>=] 0

At x = 7 or x = -7, the value of b = (x+7)(x-7) will be zero (do you know why?). Since [squ]0 is defined, f(x) is defined at these x = 7 and x = -7. At values of x between 7 and -7, the value of b will be less than 0 (verify this). That means f(x) is undefined for -7 < x < 7. For x < -7 or x > 7, b is positive (verify this also) and that means f(x) is defined.

Thus we can conclude that the natural domain of f(x) consists of all x such that

x is equal to -7, or

x is smaller than -7, or

x is equal to 7, or

x is greater than 7

We can write this as |x| [>=] 7.

Does every part of my explanation makes sense to you?

Last edited:

- #12

Lonewolf

- 337

- 1

Yes, I did mean that. Thanks.

- #13

BasketDaN

- 96

- 0

Any number that you put in for X must result in a number greater than or equal to 0 if you square that number and subtract 49. The numbers that will NOT do this are the numbers that have an absolute value less than 7.

Last edited:

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