Question

Find the domain and range of the function defined by, f(x) = 1/√9-x²
please help me.....

Answer 1

◆◆◆◆Here is your answer◆◆◆◆

Answer 2

HELLO THERE!

First, let us find the Domain of this function.

$f(x) = \frac{1}{\sqrt{9-x^{2}}}$

Note that, The denominator should be non-zero and positive.

It should be non zero because a zero in the denominator will cause the whole function to be undefined.

It should be positive because negative number under root results in an imaginary number.

So,

9 - x² > 0

=> x² < 9

By Wavy curve method, we get the domain of x to be (-3, 3).

Finding the range is a bit problematic.

The given function is an even function, so it is symmetrical about the Y axis. Now, maximum value of the term √9-x² is when x = 0. So, f(x) should be minimum at x = 0 (since reciprocal of the maximum results in minimum).

So, minimum value of f(x) is f(0), which is:

$f(0) = \frac{1}{\sqrt{9 - (0)^{2}}} \\= \frac{1}{3}$

The function is continuous, so the maximum value will be ∞.

Hence, the range is [1/3, ∞).

HOPE MY ANSWER IS SATISFACTORY...

Thanks!