find domain and rangef(X)=1/{sqrt(5)-x}

Since $\sqrt{5-x}$ is in the denominator, 5-x should be greater than zero. Otherwise the function is not defined.
So 5-x>0
5>x or x<5
So domain = $\left(-\infty ,5\right)$
Since the values of y are all greater than zero, Range =$\left(0,\infty \right)$

• 5

f(x)=1/√x-5

√x-5≥0

=>x-5≥0 =>x≥5

Domain=[5,infinity)

y= 1/ √x-5

ysquare=1/x-5

y=[0,infinity)

• -24

The above answer is wrong as sqrt.(x-5) cannot be equal to zero because then the queation will be undefined

Domain = (5,infinity)

And again the equation cannot be equal to zero as the numerator cannot be zero for any value of x. The range is from = (0,infinity)

• 3
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