shouldnt it be -5?

-5<= [2x] =< 3
and 0 <= [x2] <= 9??

Choose the following values of x

[table3b= -2, -1.5, -1, -0.5, 0, 0.5, 1][/table3b]

Accordingly the values of 2x will be

[table3b= -4, -3, -2, -1, 0, 1, 2][/table3b]

Accordingly the values of x^2

[table3b= 4, 2.25, 1, 0.25, 0, 0.25, 1][/table3b]

Accordingly the value of [2x] + [x^2]

[table3b= 0, 0, -1, 0, 0, 2, 3][/table3b]

Therefore the least value is -1

Another way to solve this problem without using derivatives is:

We have this:
\(-2.5 < x< 1.5\)
Question says that \([x]\) is the least integer grater than or equal to x
So if we think in x=-2.25 for example, it'll be rounded to -2. And x=1.25 will be rounded to 2.
As a result we'll have that
\(-2 =< [x]=< 2\) - Considering that, because of the question, in this interval we'll just have integers and we'll just focus in the endpoints, not in
the terms between,

Now we need to see the least possible value of \(2[x] + [x]^2\)
In that way:
\(-2 =< [x]=< 2\) __________Sum by 1
\(-1 =< [x]+1=< 3\) ________Squaring the interval
\(0 =< [x]^2+2[x]+1=< 9\)___Substracting by 1
\(-1 =< [x]^2+2[x]=< 8\)

So the least possible value is -1.

Using the derivative is also possible in this problem, because coincidentally the critical point "-1" is in the interval the problem gives.

But if interval is changed to \(-0.5 < x< 3\) or to \(4.3 < x< 9.6\). It won't be useful.

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 minimum


Love 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.

Please why is the square of -1 zero?

I took a different approach and will be happy to hear your opinion ifnit legit way.

-2.5<x<1.5

Doing some calculations to adjust the number to what I need.
Multiple by 2:
-5<2x<3
Square:
2.25<x^2<6.25

From that we can get all possible numbers for [2x] and [x^2].

[2x] => -4,-3,-2,-1,0,1,2,3,4
[X^2] => 3,4,5,6,7

The smallest number that we can get by adding [2x] and [X^2] is -1 because (-4+3).

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