How do you find the domain of #g(x)= x/(x^2-5x)#?

Question

Kennedy Adams · Accepted Answer

#x in RR, x != 0,5#

The domain is the range of #x# values a function can take.

The numerator of #g(x)# is simply #x# so it doesn't really matter what it is.

The denominator is more complex because we know that we can never divide by #0#.

If the denominator is #0#, then

#x^2 - 5x = 0#

#x(x-5) = 0#

thus

#x = 0, 5#

In which case the domain is #x in RR, x != 0, 5#

Isaiah Anthony · Answer

undefined The domain of $ g(x) = \frac{x}{x^2 - 5x} $ is $ x \in \mathbb{R} $ such that $ x \neq 0 $ and $ x \neq 5 $.

Jameson Allen · Answer

undefined To find the domain of $ g(x) = \frac{x}{x^2 - 5x} $, we need to identify any values of $ x $ that would make the denominator zero, as division by zero is undefined.

The denominator $ x^2 - 5x $ can be factored as $ x(x - 5) $.

For the function to be defined, the denominator cannot equal zero. So, we set $ x(x - 5) $ not equal to zero and solve for $ x $:

$$ x(x - 5) \neq 0 $$

This equation is true for all real numbers except where $ x = 0 $ and $ x - 5 = 0 $ (which means $ x = 5 $).

Therefore, the domain of $ g(x) $ is all real numbers except $ x = 0 $ and $ x = 5 $.