Question

Find a range of 1\root x^2 -9

Answer 1

Answer:

Range = R-{0}

Step-by-step explanation:

One way to solve this question is to equate the function to y, and then find an expression in terms of y and then finding the domain of the new function, which will automatically translate to the Range of the initial function.

given;

f(x)= 1/x²-9

let y=f(x);

y = 1/x²-9

y(x²-9) = 1

yx²-9y = 1

x² = (1+ 9y)/y

therefore, x = [ ( 1 + 9y)/y ]^½

thus, we see that x€ R when y is not equal to 0.

Thus Range(f(x)) = R-{0}.

Hope it helps

Answer 2

Given: 1 \ √(x² -9)

To find: The range of 1 \ √(x² -9).

Solution:

Let the given term be equal to y. To find the range of the term, the equation must be rearranged and rewritten for x in terms of y.

$y = \frac{1}{\sqrt{x^{2} - 9} }$

$y^{2} = \frac{1}{x^{2} - 9}$

$x^{2} y^{2} - 9y^{2} = 1$

$x^{2} = \frac{1 + 9y^{2}}{y^{2}}$

The denominator is y² which cannot be zero or negative. It cannot be zero since its value is equal to the term given in the question. It cannot be negative because it is a square and square of a number can only be positive. So, the range of the term is greater than zero till infinity.

Range = (0, ∞)

Therefore, the range of 1 \ √(x² -9) is (0, ∞).