please give me the domain & range of this function

Question

f(x)= 1/{sqrt(4-x^2)}.

Answer 1

Domain :

The function is f(x) = 1/{sqrt(4 - x²)}.

Domain of a function f(x) is set of those values of x which will make the function mathematically legal or correct..certain operations like division by zero , square root of a negative number do not exist in real maths.

So they are prohibited .so we have to look carefully for such possibilities to determine the domain of the function..here we have both the operations present.

1. Domain excludes x - values that result in division by zero.

The denominator expression is √(4 - x²) and it is equals to zero.

√(4 - x²) = 0

(4 - x²) = 0

x² = 4

x = ± 2.

The denominator is zero, when x = ± 2.

The function has an implied domain that consists of all real x other than x = ± 2.This value is excluded from the domain because division by zero is undefined.

2. Domain excludes x - values that result in even roots of negative numbers.

The denominator expression is √(4 - x²) and (4 - x²) should be zero or positive..zero is already prohibited above ..so it can be only positive.

=> 4 - x² > 0

=> 4 > x²

=> x² < 4

=> |x | < √4

=> |x | < 2

=> x < 2 and x > - 2.

=> The domain is all values of x such that x < 2 and x > - 2..that is x should lie between - 2 and + 2.

The notation form of domain is {x є R : - 2 < x < 2}.

Range :

Now range of f(x) is the corresponding values f(x) will take when x takes different values in the specified domain.

We find that as x varies between - 2 to 0 and from 0 to + 2 .

The expression √(4 - x²) varies between infinitely large number to 2 and again from 2 to infinitely large number.

So f(x) = 1/√(4 - x²) varies between 1 / (infinitely large number ) to 1/2 that is zero , but not equal to zero , to 1/2 so range is f(x) > 0 and f(x) ≥ 1/2.

The notation form of range is {f є R : 2 f ≥ 1}.