what is the range of the function shown below f(x)=-12/x^2+6
all possible y values over the domain
\[f(x)=\frac{-12}{x^2+6}\] if \(x=0\) then you get \(-2\) if not it is going to be larger
it cannot ever be \\(0\) because the numerator is \(-12\) so it must be \[[-2,0)\]
so is A?
\[\lim_{x \rightarrow \infty} \frac{ -12 }{ x^2 +6 } = 0\]
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The domain is all real numbers, the range is the possible y values
[-2, 0)
A
You see as you put in larger and larger numbers for x,(positive or negative), the denominator of the function becomes very large. So if the denominator becomes very large for the function, the functions value becomes very small for ever increasing size of x.
The smallest denominator yields the most negative value for f(x), which happens when you let x = 0, f(x) = -12/(0+6) = -2 The function stays less than zero, but greater than or equal to -2.
If you tested numbers near -2, you would see that the function is decreasing on the left and increasing on the right of -2, meaning -2 is the Minimum value for f(x)
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