Question

find the domain of the function algebraicly. : f(x)= square root (x^2+4) . can someone please explain to me how to do this ?

Answer 1

The domain (i.e. possible values of x) of a polynomial is all real, i.e. (-inf,+inf).
However, the square-root function sometimes puts a limit to what values x can take, in such a way that the expression within the square-root radical must be positive.
In the given problem, the expression in the sqrt radical is positive for all values of x, so the domain for f(x) is (-inf,+inf), or all real.

Answer 2

thank you so much ! i actually understand this now lol

Answer 3

You're welcome!

Answer 4

what about f(X)= 3x-1/(x-3)(x-1)

Answer 5

Cannot answer until I know what f(x) looks like. Sometimes parentheses are missed out which makes a different function!
The above function as is is the same as:
\[f(x)=\frac{(3x-1)(x-1)}{(x-3)}\]
But if missing parentheses are put around both numerator and denominator, f(X)= (3x-1)/((x-3)(x-1))it means:
\[f(x) = \frac{(3x-1)}{(x-3)(x-1)}\]

Answer 6

oh its the second one

Answer 7

In any case, this is similar to the square-root case, except that here what we don't want to see is that the denominator becomes zero.
In the first interpretation, the rational function has a domain of all real, except when the denominator becomes zero, which happens when x=3. So the domain is
(-inf,3)U(3,+inf).
This interval notation excludes 3 from the domain. "(" means exclude the value, while [ or ] includes the value.
Sometime this is also written as (-inf,+inf)\{3} which means all real excluding 3.

The answer to the second interpretation can be found in a similar way.

Answer 8

how would you do it algebraically ?

Answer 9

You can actually read it off the second interpretation by considering when the denominator will be zero.

Answer 10

It would be all real except for two values which make the denominator vanish.

Answer 11

ohh thanks your the bestt

Answer 12

You're welcome! :)