Let \(f(x) = \left\lceil\dfrac{1}{x+2}\right\rceil\) for \(x>-2\), and \(f(x) = \left\lfloor\dfrac{1}{x+2}\right\rfloor\) for \(x<-2\). (x is not defined at -2.) Which integer is not in the range of \(f(x)\)?

Really stuck with this :\ thanks for helping!

dolphinia Aug 16, 2023

#1**+1 **

Let's analyze the given function f(x) = ceiling(1/(x + 2)) for x > -2 and f(x) = floor(1/(x + 2)) for x < -2.

For x > -2: The function ceiling(1/(x + 2)) rounds up the value of 1/(x + 2) to the nearest integer. As x approaches -2 from the right (x > -2), 1/(x + 2) becomes extremely large (approaching infinity), and the ceiling of such a large positive value will also be extremely large. So, as x gets closer to -2 from the right, the function value approaches positive infinity.

For x < -2: The function floor(1/(x + 2)) rounds down the value of 1/(x + 2) to the nearest integer. As x moves further away from -2 to the left (x < -2), 1/(x + 2) becomes increasingly small (approaching 0), and the floor of such a small positive value will be 0. So, as x gets farther away from -2 to the left, the function value remains 0.

Now, let's consider the range of the function f(x):

For x > -2: The function value approaches positive infinity. For x < -2: The function value is always 0.

In other words, the range of the function f(x) consists of all positive integers and 0.

However, there is one integer missing from this range: the integer -1.

Therefore, the integer -1 is not in the range of the function f(x).

Guest Aug 16, 2023