Thread: Greatest integer function (=step function) problem

1. Greatest integer function (=step function) problem

In this question it tells me to find the zeros which means that f(x) = 0. However my problem is that, I do not know how to undo the [] brackets. Lets say we were given the following square root function and we wanted to find its zero:

f(x) = sqrt(x)
0 = sqrt(x)
0^2 = (sqrt(x))^2
0 = x
x = 0
(notice how the square root was undone by squaring sqrt(x))

for the step function:

f(x) = 3[-(x-1)/2] + 6
0 = 3[-(x-1)/2] + 6
-6/3 = [-(x-1)/2]
-2 = [-(x-1)/2]

^^ Here is where I am stuck since I do not know how to undo the [] brackets.^^

Any help would be greatly appreciated!

2. Clearly you need $\left[ { - \frac{{x - 1}}{2}} \right] = - 2$.
Do you understand that $\left[ { - \frac{3}{2}} \right] = - 2?$
Also $\left[ { -{2}} \right] = - 2$.

3. I know that [-1.5] = -2 and that [-2] = -2 however, can you spot my mistake with the first thing you mentionned?:

[-(x-1)/2] = -2
[-(x-1)] = -4
[x-1] = 4
[x] = 5

That's the furthest I can go. is that even right? The answer seems so off compared to the ]3, 5] that I am supposed to get!

4. Originally Posted by s3a
I know that [-1.5] = -2 and that [-2] = -2 however, can you spot my mistake with the first thing you mentionned?:

[-(x-1)/2] = -2
[-(x-1)] = -4

[x-1] = 4
[x] = 5
That is easy to do.
The floor function, greatest integer function, does work that way.

5. Originally Posted by s3a
The answer seems so off compared to the ]3, 5] that I am supposed to get!
Here is how it works.
$\begin{gathered}
- 2 \leqslant - \frac{{x - 1}}
{2} < - 1 \hfill \\
- 4 \le - x + 1 < - 2 \hfill \\
3 < x \leqslant 5 \hfill \\
\end{gathered}$
.

6. Originally Posted by s3a
I know that [-1.5] = -2 and that [-2] = -2 however, can you spot my mistake with the first thing you mentionned?:

[-(x-1)/2] = -2
[-(x-1)] = -4
[x-1] = 4
[x] = 5

That's the furthest I can go. is that even right? The answer seems so off compared to the ]3, 5] that I am supposed to get!
[x]= n, for n an integer, is true for any x between n and n+1.
It is NOT true that if [-(x-1)/2]= -2 then [-(x-1)]= -4. For example if x= 3.5, then -(x-1)= -(2.5)= -2.5. so -(x-1)/2= -1.25 so [-(x-1)/2]= -2. But [-(x-)]= -3, not -4.

Better to think: if [y]= -2 then $-2\le y< -1$. Now replace y by -(x-1)/2: $-2\le -(x-1)/2< -1$ so $-4\le -x+1< -2$, $-5\le -x< -3$, $3< x\le 5$

Blast you, Plato! Again, 2 minutes ahead of me!