Help with 3!

3. This is a "floor" function.....it returns the greatest integer less than or equal to f(x)

f(1)  =  floor ( -1/6)  = -1

f(1000)  =  (-2998/1005)  ≈ -2.98   ...so the floor of this = -3

We need to find the "break" points

We need to first determine what integer value [approximately]  makes  (2 -3x) / ( x + 5)  = -1

So   we have

(2-3x) / (x + 5)  = -1

(2 - 3x)  = -1(x + 5)

2 -3x =  -x -5

7 = 2x

3.5  = x

Note that  f(3)  produces  floor (-7/ 8)  = -1

And f(4)  produces floor (-10/9)  = -2

Next.....we need to determine what integer value [ approximately] makes (2- 3x) / ( x + 5)  = -2

So we have

(2 - 3x) / ( x + 5)  = -2

2 - 3x  = -2(x + 5)

2 -3x  = -2x -10

12 = x 

Note that f(12) produces  floor (-34/17)  = -2

And f(13)  produces  floor (-37/ 18)  = -3

So...from f(1)  to f(3)...we will  have  the sum 3(-1)  = -3

And from f(4)  to f(12)  we will have the sum  9(-2)  = -18

And from f(13) to f(1000)  we will have the sum  988(-3)  = -2964

So....the sum of the series  =   (-3 ) + ( -18 ) + ( -2964 )  = -2985

This graph shows the break points :

https://www.desmos.com/calculator/q95fsnkc0t

The equation \(\left \lfloor{t}\right \rfloor=2t+3\) can be a tad difficult to parse. Sometimes, it is better to think of an easier problem before attempting one a harder one. Let's think about a basic equation that contains the floor function.

\(\left \lfloor{t}\right \rfloor=5\)

Before we can solve this, though, we have to understand the floor function. The floor function truncates any decimal. The solution set for this simplified equation, represented in a compound inequality, is \(5\leq t<6\). Any value for t that satisfies this restriction is a solution. We can use this logic to generalize the floor function, actually to \(\left \lfloor{x}\right \rfloor=a\rightarrow a\leq x < a+1\) . However, you must have caution when doing this, however. Not every solution is actually a solution. I'll touch on this later. Knowing this will be key to solving this equation. Knowing this, the floor function transforms from  \(\left \lfloor{t}\right \rfloor=2t+3\) to  \(2t+3\leq t<2t+3+1\). Let's solve this compound inequality. I'll solve each one separately.

Now, let's solve for the other inequality \(t<2t+3+1\):

Of course, all solutions must adhere to both restrictions. Both inequalities show that t must be greater than -4 and less than or equal to -3. -3 is automatically a solution because it is an integer.

We aren't done yet, though. Are there are other solution that adhere to our current restriction that makes \(2t+3\) an integer? We don't even have to worry about the +3 because the sum of any integer and 3 will always yield another integer. Let's set this equal to 0:
 

What I have demonstrate here is that a number divided by 2 will make 2t+3 an integer. The only one that meets our description and our current restriction is -(7/2).

Therefore, there are 2 solutions to this equation

\(t_1=-\frac{7}{2}\)

\(t_2=-3\)

Sorry, X²......I think you made a slight mistake.....your solution of -3 is correct, but not -3/2

Floor (-3/2)  = -2

But

2(-3/2) + 3  =  0

The other solution is   t  = -7/2

Floor (-7/2)  = Floor (-3.5)  =  -4

And

2(-3.5) + 3  =

-7 + 3  =   -4

Here's a graph of the solution points :