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#1 2007-10-13 01:27:37

Identity
Member
Registered: 2007-04-18
Posts: 934

Domain and Range

Find "algebraically" the domain and range of

. How could I do this? Thanks

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#2 2007-10-13 02:31:01

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: Domain and Range

Domain: Basically you want to see what values of x can go into the function and produce some sort of valid output. Assess things like whether there are any values of x where the denominator is zero, or where the denominator doesn't exist (I assume you're working with the real numbers here, right?)

Range: Find out if there are any maximum or minimum values that will prevent the range from being all the real numbers.

Remember that just because it says you have to find them algebraicly, that doesn't mean you can't draw a graph to give you some direction.


Bad speling makes me [sic]

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#3 2007-10-13 03:02:42

Identity
Member
Registered: 2007-04-18
Posts: 934

Re: Domain and Range

Is this how you would do the domain?

We need for

So...

or

So x is positive whenever

or
So the domain is
.

For the range... is there any easy way to do it other than differentiating it? I'm not supposed to apply calculus to it.

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#4 2007-10-17 07:04:36

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Domain and Range

You don't need to. As x --> ∞, x² - 6x --> ∞ also.
x² - 6x can go as high as it wants, and therefore f(x) can get as close to 0 as it wants.

Similarly, as x --> 6+, x² - 6x --> 0.

x² - 6x can get as close to 0 as it wants, so f(x) can go as high as it wants.

Clearly f(x) can never be negative since 1 and √(x² - 6x) are both always positive.

Therefore, your range is (0,∞).

(I've used limits there, which are technically calculus, but you could get away with it if you presented the arguments as only intuitive).


Why did the vector cross the road?
It wanted to be normal.

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