Jon P. answered • 08/07/15

Knowledgeable Math, Science, SAT, ACT tutor - Harvard honors grad

Gautam K.

asked • 08/07/15chapter = relations and functions

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

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Jon P. answered • 08/07/15

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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.

Ketaki M. answered • 08/07/15

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Ketaki M Math,SAT,ACT.GRE,AP CAlculus

for the domain of the above function,9-x^{2 }should be greater than 0.

ie x^{2}-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-x^{2 }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

Michael J. answered • 08/07/15

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f(x) = 1 / √(9 - x^{2})

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 - x^{2} > 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 - 4^{2})

= 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 - 0^{2})

= 1 / 3

= 0.33

f(1) = 1 / √(9 - 1^{2})

= 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.

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