# h(x) =( 5-x)/the square root of 5x-125what is largest possible domain

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The denominator of the function impose the domain of definition.

We'll impose the constraint of existence of the function:

5x-125 > 0

We've imposed that the expression to be strictly positive, since the sqrt 5x-125 represents the denominator. According to the rule, the values of denominator have to be other than zero.

We'll divide by 5:

x - 25>0

We'll add 25 both sides:

x > 25

**The domain of definition is the interval of x values, that are strictly greater than 25!**

**x belongs to (25, +infinite)**

To find the largest domain of h(x) = (5-x)/sqrt(5x-125)

The lthe domain of h(x) = (5-x)/sqrt(5x-125) is the set of all x for which h(x) is defined and real.

The expression (5-x)/ sqrt(5x-125) has the denominator sqrt(5x-125).

When 5x-125 = 0, x = 125/5 = 25.

So the expresion (5-x)/sqrt(5x-125) becomes (5-25)/0 which is not defined.

So x = 25 cannot be in the domain.....(1)

Also sqrt(5x-125) is not real when 5x < 125. Or x < 125/5 = 25.

Therefore the domain cannot take vaues x < 25. (2)

So combining the (1) and (2) , we see that the domain of h(x) = (5-x)/sqrt(5x-125) is x > 25. Or the domain of x is (25 , infinity) wher 25 is not included.

The domain of a function is defined as the values of x for which the function takes real values.

Here the function we have is g(x) = ( 5-x)/ sqrt (5x-125)

Now g(x) has real values for all values of x except those where the term we are finding the square root of, is negative. Also, as sqrt (5x-125) is in the denominator it cannot be equal to 0. So only values of x where 5x-125 > 0 are part of the domain.

5x - 125 > 0

=> 5 ( x- 25 ) >0

=> x -25 >0

=> x > 25

Therefore the function g(x) = ( 5-x)/ sqrt (5x-125) gives real values where x > 25.

**The domain is all values of x greater than 25.**