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interval function

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2 Answers

Is the denominator of the function (x2-8), or is it x2?
 
As written, with parenthesis for clarification, the function is:
f(x) = (1/x2) - 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/x2), 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/(x2-8), we will again consider where the denominator may equal 0, and exclude these values from the domain.
 
x2 - 8 = 0
x2 = 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.
 
 
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;∞).