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#1 2022-09-25 10:47:59

sologuitar
Member
Registered: 2022-09-19
Posts: 467

Find Domain of f(x)

Find the domain of f(x) = 5/[sqrt{x^2 + 1}].

Again, the domain is ALL REAL NUMBERS but it is not clear to me why that is the case.

According to Professor Leonard on YouTube, for this function, no matter what real numbers we let x be, the radicand cannot yield a negative number.
So, there is no domain issue here. The domain is ALL REAL NUMBERS.
However, I need a more detailed reason why the domain is ALL REAL NUMBERS. Can a graph of this function help me to see why Professor Leonard is right?

Thanks

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#2 2022-10-30 10:26:42

mathdrop
Member
Registered: 2022-03-07
Posts: 75

Re: Find Domain of f(x)

You could look at the function graph and see if it helps you ;-)

Now let's look at the equation:

x² does always yield positives. even if you set x to a negative.
Could this already be the reason ?

Further its always even and x²+1 is always uneven.

Taking the root of positives gives you real numbers
(as opposed to imaginary numbers if you take it from a negative).

Cause of the +1,
1 would be the smallest number, you would have in the root.

1 times 1 is 1, so the square root of 1 equals 1.
5/1 equals 5.
Since there is no shifting, the numerator,
in this case 5, is the y-Intercept here.

So much for now.

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#3 2022-10-30 20:25:21

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Find Domain of f(x)

The default is all reals. This is only reduced if a questioner chooses or if there are values of X that make y impossible to calculate.

A square root might cause this if we are required to find the root of a negative. You have correctly identified that this doesn't happen here so no need to limit the domain at all.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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