what is the range of f(x) = 1/(9-x^2)^1/2

Question

chapter = relations and functions

'^' stands for 'raised to ' or 'to the power'

Answer 1

If x is 0, f(x) = 1/√9 = 1/3.

 

As x increases, the denominator gets smaller, so f(x) gets larger.  The function continues to grow "without bound" as x approaches 3, and when x is very close to 3, the function is very large.  This continues until x = 3, at which point the denominator is undefined. 

 

When x > 3, the expression inside the square root is negative.  I'm guessing that you're not allowing for complex numbers, so let's say that when x > 3, the function is also undefined.

 

Now as x decreases from 0, the behavior of the function is the same -- it grows very large until x = -3, at which point it becomes undefined.

 

So the range is all real numbers greater than or equal to 1/3.

 

 

Answer 2

for the domain of the above function,9-x²  
should be greater than 0.

 

ie x²-9<0⇒(x-3)(x+3) <0

 

which by the wavy curve method would give you  x∈(-3,3)

 

now since it is an even function,the values of y can be found from the mid point of the domain ,to one extremity of it.

 

so we will find the value of y at x=0 and we get y as 1/3. this will be an attainable value since x=0is a closed value in the domain.

 

now we will think about when x tends to 3.as x tends to 3,√9-x² tends to 0.

 

so y tends to 1/0 

 

which is infinite

 

so the range of given function is [ 1/3,∞) I would suggest you use a graphing calculator to see it

Answer 3

f(x) = 1 / √(9 - x²)

 

 

The range is the set of y values in which the graph exist.  In order to find the range we can find the domain first. Then we can evaluate f(x) using those set of x values in the domain.

 

9 - x² > 0

 

(3 - x)(3 + x) > 0

 

 

If x=-3  and   x=3 are our zeros, then the values that make the statement positive using test points are

 

 

f(-4) = 1 / √(9 - (-4)²)

        = 1 / √-16

 

 

f(-2) = 1 / √(9 - (-2)²)

        = 1 / √5

 

f(4) = 1 / √(9 - 4²)

       = 1 / √-16

 

 

Domain is (-3 , 3).

 

Now we evaluate f(x) using some x-values in the domain.

 

f(-2.9) = 1 / √(9 - (-2.9)²)

           = 1 / √0.59

           = 1.3

 

f(-1) = 1 / √(9 - (-1)²)

        = 1 / √8

        = 0.35

 

f(0) = 1 / √(9 - 0²)

       = 1 / 3

       = 0.33

 

f(1) = 1 / √(9 - 1²)

       = 0.35

 

Note that f(-1) and f(1) are the same.  This means that f(x) gets closer to a minimum value of 0.33.   f(x) gets closer to a maximum value of 1.3.

 

 

The range is  (0.33 , 1.3).

 

 

Feel free to use a graphing calculator to verify these values.