Carcass wrote:
For all values, \([x]\) denotes the least integer greater than or equal to x. If \(-2.5 < x< 1.5\), what is the least possible value of \([2x] + [x^2]\)?
Notice that when \(x = 0\) the function is at 0, and when we start to hit positive numbers in range of \(x\) that the numbers will be greater than this minimum we've found of 0.
We might guess that if there is a minimum below 0, it must be in the negative range of \(x\) since \([2x]\) can be negative if \(x\) is negative. So let's start from the bottom end of the range \(-2.5 < x < 1.5\) and work our way up.
\(x = -2.4\)
\([2(-2.4)] + [(-2.4)^2]\)
\([-4.8] + [5.76]\)
These numbers go towards the closest greatest integer, so \([-4.8]\) goes to -4, and \([5.76]\) goes to 6. \(-4 + 6 = 2\).
Not quite the minimum we're looking for, since by observation we can get 0 if \(x=0\). Let's keep working our way up:
\(x = -2\)
\([2(-2)] + [(-2)^2]\)
\([-4] + [4]\)
\(0\)
We've arrived at 0. Let's try again.
\(x = -1.5\)
\([2(-1.5)] + [(-1.5)^2]\)
\([-3] + [2.25]\)
\(-3 + 2 = -1\)
And there we find -1. In fact, any number in the range \(-2 < x < 0\) will give us the minimum.
From above, we know that the positive numbers will give us positive integer results, so we don't have to do those calculations.
Therefore, -1 is our minimumLove the answers using calculus and building from the inequality! Though I would abandon the use of calculus for this exam, as it might lead you down slippery slopes in other problems.