Domain: the 5x + 2 is in squareroot form as the denominator.

[violeti]

The problem is 5x + 2. The 5x + 2 is in squareroot form as the denominator.

 

[Aladdin]

Re: Domain

The problem is 5x + 2. The 5x + 2 is in squareroot form as the denominator.

5x+2 >0 Since in real numbers under the radical must alwasy be positive? right ?
5x>-2
x>-2/5

 

[stapel]

5x+2 >0 Since in real numbers under the radical must alwasy be positive? right ?

No; the argument of a square root can be zero.

The problem is 5x + 2. The 5x + 2 is in squareroot form as the denominator.

I will guess that you mean that you have some function along the lines of the following:

. . . . .\(\displaystyle f(x)\, =\, \frac{\mbox{(something)}}{\sqrt{5x\, +\, 2}}\)

...and you have been asked to find the domain of the function...?

If so, then note that there are two restrictions here. You cannot have a negative inside the square root, and you cannot have zero in the denominator. So you must solve \(\displaystyle 5x\, +\, 2\, \ge\, 0\) for the square root.

In addition, you must also specify that \(\displaystyle \sqrt{5x\, +\, 2}\, \neq\, 0\). Together, these two restrictions give you the one combined restriction of \(\displaystyle 5x\, +\, 2\, >\, 0\).

Make sure you understand how this restriction arose, so that you can correctly interpret other exercise. :wink:

 

[Aladdin]

Stapel, cant you directly say : 5x+2>0 , without justification (( I mean in the test)) ? ?

 

[stapel]

Stapel, cant you directly say : 5x+2>0 , without justification (( I mean in the test)) ? ?

I'm not sure what you're asking...?

But, in general, no, one cannot say that a square root can never have an argument of zero, because the square root of zero is well-defined; namely, the square root of zero is zero. In the particular circumstances of this specific function, it is a different issue that places the added restriction on the argument of the radical. In order that the student not be misled, your apparently invalid reasoning needed to be corrected.

 

[Aladdin]

Re:

Stapel, cant you directly say : 5x+2>0 , without justification (( I mean in the test)) ? ?

I'm not sure what you're asking...?

But, in general, no, one cannot say that a square root can never have an argument of zero, because the square root of zero is well-defined; namely, the square root of zero is zero. In the particular circumstances of this specific function, it is a different issue that places the added restriction on the argument of the radical. In order that the student not be misled, your apparently invalid reasoning needed to be corrected.

Ohh okay. Thank you

 

[Deleted member 4993]

Stapel, cant you directly say : 5x+2>0 , without justification (( I mean in the test)) ? ?

Yes you can - but you said

5x+2 >0 Since in real numbers under the radical must alwasy be positive? right ?

That would be wrong logic.

 

[Aladdin]

Stapel, cant you directly say : 5x+2>0 , without justification (( I mean in the test)) ? ?

Yes you can - but you said

5x+2 >0 Since in real numbers under the radical must alwasy be positive? right ?

That would be wrong logic.

Yes, but I'm right . In real numbers its impossible to have a negative sigh under the radical I mean the answer.

 

[Deleted member 4993]

No - you are not ....

"Not positive" -does not mean "negative".

 

[Aladdin]

No - you are not ....

"Not positive" -does not mean "negative".

Thank You Khan, you're right.