For what values of x are these functions defined?

[Xonian]

Hey guys. I was doing my math homework and ran into some trouble with numbers 42 and 43, both are attached in a picture. I could find the rational functions just fine, but I did not quite understand part b. What does it mean for a function to be defined? The answer for part b of number 43 was all real numbers greater than zero, but I don't understand why? Here is what I got for part a for both numbers:

42.

a. R(x)= (6x+4)/(2x^2 +3x+1)


43.

a. R(x)= 2/x

If you could please help, that would be great. The two math problems have been attached.

 

[stapel]

Can you have negative values for side lengths, circumferences, perimeters, area, etc?

 

[Xonian]

Can you have negative values for side lengths, circumferences, perimeters, area, etc?

That's all defined means? Haha, thanks.

 

[Quaid]

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That's all defined means?

No, but you need to consider the context of the exercises.

For example, the expression 2/x is defined for all Real numbers x, except zero. That is, numbers like 2/(-10) make sense mathematically, but x=-10 does not make sense in the context of the exercise. That's why the answer is all positive Real numbers.

Post your answers, for all of the restricted domains, and we'll check 'em.

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[Xonian]

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No, but you need to consider the context of the exercises.

For example, the expression 2/x is defined for all Real numbers x, except zero. That is, numbers like 2/(-10) make sense mathematically, but x=-10 does not make sense in the context of the exercise. That's why the answer is all positive Real numbers.

Post your answers, for all of the restricted domains, and we'll check 'em.

ֺ

Ok, for 42 part b, I got all real numbers above -2/3 for P, all reals above -1/2 for A, and all reals above -1/2 for R. For 43 part b, I got all reals above zero for all three.

And one quick question. For number 42, when it is asking for what values of x define P, or 6x+4, the numerator, do I just look at that polynomial, or do I look at the denominator too, considering that is a separate function. Thanks!

 

[Quaid]

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For 43 part b, I got all reals above zero for all three.

That's correct.

for 42 part b, I got all real numbers above -2/3 for P

all reals above -1/2 for A, and all reals above -1/2 for R

Check your restricted domain for P.

EG: -3/5 is greater than -2/3, but the width of the rectangle (2x+1) is negative for x = -3/5

You're looking for the smallest value of x, such that both the width and the height are positive numbers.

when it is asking for what values of x define P, or 6x+4, the numerator, do I just look at that polynomial, or do I look at the denominator too, considering that is a separate function.

Are you confusing R with P?

P does not have a denominator; it's simply 6x+4. So, if you're asked about P, you don't consider A=(2x+1)(x+1).

If you're asked about R, then you need to consider R=P/A -- that is, the ratio (6x+4)/((2x+1)(x+1)).

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[Xonian]

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Check your restricted domain for P.

EG: -3/5 is greater than -2/3, but the width of the rectangle (2x+1) is negative for x = -3/5

You're looking for the smallest value of x, such that both the width and the height are positive numbers.


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Ok, everything makes sense except this. When P, the perimeter, is simplified, it equals 6x+4, derived from 2(x+1)+2(2x+1). So for that to be defined, wouldn't the x values have to be greater than -2/3? If you plug in -2/3 you get zero. Any higher number results in a positive corresponding y-value.

 

[Quaid]

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This is where you get to expand your thinking, a bit.

Yes, it is true that values of x greater than -2/3 result in positive values of 6x+4, but you're not asked for values of x that make P positive. You're asked about values of x that make sense, in the context of the exercise.

We need to remain cognizant of where 6x+4 comes from. This perimeter expression comes from adding the four lengths. Those lengths must all be positive.

When x = -3/5, the rectangle's height (x+1) is positive, but its width (2x+1) is not. In other words, there are some values greater than -2/3 which do not work.

You need to be sure that all values of x in your restricted domain lead to positive measurements for each of the rectangle's two dimensions.

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[Xonian]

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This is where you get to expand your thinking, a bit.

Yes, it is true that values of x greater than -2/3 result in positive values of 6x+4, but you're not asked for values of x that make P positive. You're asked about values of x that make sense, in the context of the exercise.

We need to remain cognizant of where 6x+4 comes from. This perimeter expression comes from adding the four lengths. Those lengths must all be positive.

When x = -3/5, the rectangle's height (x+1) is positive, but its width (2x+1) is not. In other words, there are some values greater than -2/3 which do not work.

You need to be sure that all values of x in your restricted domain lead to positive measurements for each of the rectangle's two dimensions.

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Awesome, thank you!