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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.
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