give the range and the domain in

**interval notation**of f(x)= 1/x^2-8give the range and the domain in **interval notation** of f(x)= 1/x^2-8

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Athens, GA

Is the denominator of the function (x^{2}-8), or is it x^{2}?

As written, with parenthesis for clarification, the function is:

f(x) = (1/x^{2}) - 8

The domain of the function is the set of real number values that we may input into the function and get a real number value as an output. Since division by 0 is undefined, the domain of the function (as written) is all real numbers except 0.

If we compare this function to (1/x^{2}), which has a range of (0, infinity). Now the function that you wrote has been vertically shifted down 8 units, and the range would be (-8,infinity)

If, instead, you meant:

f(x) = 1/(x^{2}-8), we will again consider where the denominator may equal 0, and exclude these values from the domain.

x^{2} - 8 = 0

x^{2} = 8 (taking the square root of both side and evaluating the result)

x = ±2√(2)

Then the function has two vertical asymptotes at these values (x=2√(2) and x= - 2√(2)), and the domain is all real numbers except these two values.

Since the highest power of x in the numerator is less than the highest power of x in the denominator, there is a horizontal asymptote at y=0.

The range of this function is all real numbers not equal to 0.

Willowbrook, IL

For domain:

The only special points are those when denominator is zero. Then the function is not defined and those points shall be excluded from domain.

x2-8=0;

x=±2√2;

Thus domain, D, of a function is D:x∈(-∞;-2√2)∪(-2√2;2√2)∪(2√2;∞).

For the range:

Consider the denominator. Since x2 is always non-negative, x2-8 is always greater or equal to -8. When the x approaches -2√2 from the right of 2√2 from the left, denominator is negative and goes to zero, therefore the whole function goes to minus infinity. So on the interval (-2√2;2√2) the function f(x)=1/(x2-8) goes from -1/8 to -∞. On two other intervals the function goes to +infinity when x approaches -2√2 from the left or 2√2 from the right. When x goes to ±∞, denominator goes to +∞ and the whole function, f(x), goes to zero, but never attains that value. Thus, the range of this function is:

R: f(x)∈(-∞;-1/8]∪(0;∞).

The only special points are those when denominator is zero. Then the function is not defined and those points shall be excluded from domain.

x2-8=0;

x=±2√2;

Thus domain, D, of a function is D:x∈(-∞;-2√2)∪(-2√2;2√2)∪(2√2;∞).

For the range:

Consider the denominator. Since x2 is always non-negative, x2-8 is always greater or equal to -8. When the x approaches -2√2 from the right of 2√2 from the left, denominator is negative and goes to zero, therefore the whole function goes to minus infinity. So on the interval (-2√2;2√2) the function f(x)=1/(x2-8) goes from -1/8 to -∞. On two other intervals the function goes to +infinity when x approaches -2√2 from the left or 2√2 from the right. When x goes to ±∞, denominator goes to +∞ and the whole function, f(x), goes to zero, but never attains that value. Thus, the range of this function is:

R: f(x)∈(-∞;-1/8]∪(0;∞).

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