Finding the domain of y = sqrt(3x + 2)

[Gadsilla]

I don't understand how to start. Also, I'm experiencing a problem in general when doing calc where it makes sense when someone explains an issue for me, but I don't know how I would do it myself if the question is switched up. Any tips ?

 

[Harry_the_cat]

I don't understand how to start. Also, I'm experiencing a problem in general when doing calc where it makes sense when someone explains an issue for me, but I don't know how I would do it myself if the question is switched up. Any tips ?

Basically, the domain is the set of values that x can take, so that the function is defined.

For the function $\displaystyle y =\sqrt{3x+2}$ to be defined, then 3x+2 >=0 (since you can't take the square root of a negative number).

Now, solve this inequality for x, and you have the domain.

 

[Gadsilla]

Basically, the domain is the set of values that x can take, so that the function is defined.

For the function $\displaystyle y =\sqrt{3x+2}$ to be defined, then 3x+2 >=0 (since you can't take the square root of a negative number).

Now, solve this inequality for x, and you have the domain.

I don't understand, where's the negative ?

 

[Steven G]

I don't understand, where's the negative ?

What is under the square root sign is the place holder for a number. Now you can not compute the square root of a negative number. You can only compute the square root of 0 or a positive number. So to compute the domain you simple set what is under the square root sign to be greater than or equal to zero and solve for x. This restriction on x will be the domain

For the record, everything above is correct if we are talking only about real numbers

 

[JeffM]

My answer is basically the same substantively as jomo's and harry's, but sometimes a different set of words helps.

Someone can explicitly define a domain such as

$\displaystyle f(x) = \sqrt{4 - x} \text { for all real } x \ge 4.$

It is not hard to figure out the domain when it is given to you. But sometimes you are given the function without any explicit definition of the domain. Then you are expected to figure out what the implied domain is. In beginning calculus, you generally are dealing with real numbers, not complex numbers or quaternions or the super reals. In that kind of problem, you are supposed to identify the largest subset of real numbers that will result in a real number as a result.

$\displaystyle f(x) = \sqrt{x^3 - 27}.$

You know that the square root of a negative number is not a real number. Under what circumstances will

$\displaystyle x^3 - 27 \ge 0$? Obviously, only if $\displaystyle x \ge 3.$ The implied domain for that function is every real number not less than 3.

$\displaystyle g(x) = \dfrac{x + 1}{x - 1}.$

You know that division by zero does not result in a real number. Under what circumstances will

$\displaystyle x - 1 \ne 0.$ Obviously, if $\displaystyle x \ne 1.$ So the domain is all real numbers except 1.

$\displaystyle h(x) = \sqrt{\dfrac{x + 1}{x - 1}}.$

This one is bit tougher. You know that division by zero does not result in a real number nor does the square root of a negative number. We cannot have

$\displaystyle x = 1$.

Nor can we have the fraction be negative.
Because x - 1 < x + 1, the fraction will not be negative if

$\displaystyle 0 \le x - 1 < x + 1 \text { or } x - 1 < x + 1 \le 0.$ To summarize,

the fraction will not be a negative if $\displaystyle 1 \le x \text { or } x \le -\ 1.$

$\displaystyle \therefore \text {The domain of } h(x) \text { is all real } x \text { such that } x \le -\ 1 \text { or } x > 1.$

 

[lookagain]

Someone can explicitly define a domain such as

$\displaystyle f(x) = \sqrt{4 - x} \text { for all real } x \ge 4$

Did you intend $\displaystyle f(x) = \sqrt{x - 4 \ } \$ instead?

$\displaystyle x - 4 \ \ge \ 0$

$\displaystyle x \ \ge \ 4$

 

[Gadsilla]

Basically, the domain is the set of values that x can take, so that the function is defined.

For the function $\displaystyle y =\sqrt{3x+2}$ to be defined, then 3x+2 >=0 (since you can't take the square root of a negative number).

Now, solve this inequality for x, and you have the domain.

I don't understand ? There's no negative number.

 

[Harry_the_cat]

I don't understand ? There's no negative number.

But $\displaystyle 3x+2$ COULD BE negative. You have to make sure that $\displaystyle 3x+2$ is NOT negative

Ok. So $\displaystyle y = \sqrt{3x+2}$.

If, for example, $\displaystyle x=4$ then $\displaystyle y=\sqrt{3*4+2} = \sqrt {14}$ which is great because you can find the square root of 14, ie the square root of 14 exists (it's not rational but it exists). Therefore $\displaystyle x=4$ is part of the domain.

If, for example,

$\displaystyle x=-4$ then $\displaystyle y=\sqrt{3*-4+2} = \sqrt {-10}$. This isn't good because you can't take the square root of -10, or any negative number for that matter (in the real number system anyway, but don't worry about that for now).

Therefore $\displaystyle x=-4$ is NOT part of the domain. Because when you plug in -4 for x, the answer you get doesn't exist.

So,
the bit underneath the square root sign can NOT be negative, in other words it has to be positive or 0.

That is:

$\displaystyle 3x+2 \geq0$. SO, solve this inequality for x and you'll have the domain.​

 

[Gadsilla]

But $\displaystyle 3x+2$ COULD BE negative. You have to make sure that $\displaystyle 3x+2$ is NOT negative

Ok. So $\displaystyle y = \sqrt{3x+2}$.

If, for example, $\displaystyle x=4$ then $\displaystyle y=\sqrt{3*4+2} = \sqrt {14}$ which is great because you can find the square root of 14, ie the square root of 14 exists (it's not rational but it exists). Therefore $\displaystyle x=4$ is part of the domain.

If, for example,

$\displaystyle x=-4$ then $\displaystyle y=\sqrt{3*-4+2} = \sqrt {-10}$. This isn't good because you can't take the square root of -10, or any negative number for that matter (in the real number system anyway, but don't worry about that for now).

Therefore $\displaystyle x=-4$ is NOT part of the domain. Because when you plug in -4 for x, the answer you get doesn't exist.

So,
the bit underneath the square root sign can NOT be negative, in other words it has to be positive or 0.

That is:

$\displaystyle 3x+2 \geq0$. SO, solve this inequality for x and you'll have the domain.​

How can you move on assuming that it COULD be a negative ? It makes more sense now, but is that what you're supposed to do everytime if (x) can be a negative ?

 

[mmm4444bot]

How can you move on assuming that it COULD be a negative ?

Are you thinking of the expression 3x+2, where you asked about 'it'?

 

[Gadsilla]

Are you thinking of the expression 3x+2, where you asked about 'it'?

Any formula that could have a negative, but yes.

 

[Harry_the_cat]

How can you move on assuming that it COULD be a negative ?

Well it (3x+2) COULD be negative, it COULD also be positive or zero. It depends on what x is. Doesn't it?
​

It makes more sense now, but is that what you're supposed to do everytime if (x) can be a negative ?

No! Not f(x), only the bit underneath the square root sign.

If f(x) involves a square root, then only the bit under the square root sign has to be non-negative for the function to be defined. That is what the domain is - all the values of x that allow the function to be defined. The square root of a negative number is not defined (for real numbers).

For example,

The function $\displaystyle f(x) = 5x^2 - 27x +32.5 +\sqrt{3x+2}$ will only be defined when $\displaystyle 3x+2 \geq0$, that is $\displaystyle x\geq \frac{-2}{3}$. And so that's the domain (all the values of x that allow the function to be defined).

So, again, if a function involves a square root, then the bit under the square root sign must be positive or zero.

Your function involved a square root. You can't take the square root of a negative number, so you have to define the domain so that that doesn't happen.

Looking at a different example , say $\displaystyle f(x) = \frac{1}{x-4}$.

No square roots this time, but this function involves division. You know that you can't divide by 0. Dividing by 0 is undefined, just like taking the square root of a negative is undefined.

So you've got to make sure that $\displaystyle x-4\neq0$. So $\displaystyle x\neq 4$.

In this case the domain is every real number except 4.

You might like to go back now and have another read of JeffM's response earlier.

 

[Steven G]

How can you move on assuming that it COULD be a negative ? It makes more sense now, but is that what you're supposed to do everytime if (x) can be a negative ?

Sometimes the domain Can include negative number. Consider the sqrt(9-x). Now if x=-7 all is ok since sqrt(9-(-7)) = sqrt(9+7)= sqrt(16) and 16 IS positive. So x=-7 IS in the domain.
To find the domain of the problem I stated you need to solve 9-x>=0 and get 9<=0. In fact, you should know that if take away more than 9 from 9, then the result will be negative which is not allowed. So x>=9.

 

[Harry_the_cat]

Sometimes the domain Can include negative number. Consider the sqrt(9-x). Now if x=-7 all is ok since sqrt(9-(-7)) = sqrt(9+7)= sqrt(16) and 16 IS positive. So x=-7 IS in the domain.
To find the domain of the problem I stated you need to solve 9-x>=0 and get 9<=0. In fact, you should know that if take away more than 9 from 9, then the result will be negative which is not allowed. So x>=9.

I think you meant 9<=x.

 

[sinx]

How can you move on assuming that it COULD be a negative ? It makes more sense now, but is that what you're supposed to do everytime if (x) can be a negative ?

domain
1/(x-1); x can't be 1 (you cannot divide by 0)
domain for x is all numbers except for 1

sqrt (x-1); x has to be >or= 1 (you cannot take sqrt of a negative number)
domain for x is all numbers >or= 1

 

[mmm4444bot]

Any formula that could have a negative, but yes.

In this case, you're thinking of a completely different situation.

If you have a function containing a negative radicand, then it's a complex-valued function (i.e., the function outputs Complex numbers, most of which are probably not Real). Such functions are not a pre-calculus topic.

You don't seem quite ready for introductory calculus. I suspect you may need to back up to algebra (or possibly pre-algebra) and review. I think you need more practice, in the prerequisite topics.

Also, your new question about domains for "any formula" sounds like you may be going off-topic. If you're concerned with seeking 'one-size-fits-all' approaches (to make things "simpler"), that could be a waste of your time, at this point. Finding a domain for some Real-valued function is kinda like a process of elimination. Start by assuming it's all Real numbers, and then use what you know about algebra rules, to eliminate values of x that cause things to "break" (i.e., violate rules of algebra). For examples, we can't divide by zero, so if your function contains an expression in the denominator of a ratio, then x cannot take on any value which causes the denominator to evaluate to zero. If your function contains a radicand, then x cannot take on any value which causes the radicand to be negative, etc. Once you've accounted for all "illegal" values of x, whatever remains is the domain.

My suggestion is that you post one concrete exercise per thread, and, if you become curious about another matter, start a new thread. :cool:

 

[mmm4444bot]

… I'm experiencing a problem in general when doing calc where it makes sense when someone explains an issue for me, but I don't know how I would do it myself … Any tips?

You haven't practiced enough. Get extra exercises; practice, practice, practice, until you reach the point where you are confident that you do know how to proceed. :cool:

 

[Gadsilla]

You haven't practiced enough. Get extra exercises; practice, practice, practice, until you reach the point where you are confident that you do know how to proceed. :cool:

Will do.

Also, I'm already taking calc 1. I'm just revisiting topics I never properly learned. I'm doing alright in calc.