How would you answer this question without absolute value notation? Ix-3I+lx-4l, x<3?

Answer 1

Ix-3I + lx-4l, x<3
since x<3
x-3 <0
x -4 < 0
so
Ix-3I + lx-4l = 3-x + 4-x = 7 - 2x

Answer 2

Well, from what you've written, there's not one answer, it's a function. We know that x is less than 3, but there's no "lower limit" on it, so it could be as large as negative infinity (what a thought!). If x=3 (highes possible value), we get 1 (3-3 plus 3-4, but no answer is allowed to be negative so 0+1=1). If x = 2 (nothing you said limited x to integers, but let's keep it simple), we have 3 (2-3 + 2-4). If x=1, it's 5. If x=0, it's 7. See the pattern: 1,3,5,7 when x is 3,2,1,0. So, if we can come up with another equation that gives us this, we're done.

And, if x=3 produces 1, and x=2 produces 3, and so on, the following equation works:

7 - 2x, where x<3

Answer 3

19

Answer 4

Find the points which are zeros of one of the absolute value:
x=3 and x=4 are such points

When x<=3 x-3<=0 and |x-3|=-(x - 3)
When x>=3 x-3>=0 and |x-3|=(x - 3)
When x<=4 x-4<=0 and |x-4|=-(x - 4)
When x>=3 x-3>=0 and |x-4|=(x - 4)

Split R into 3 intervals:
From -inf to 3
From 3 to 4
and from 4 to inf.

Answer 5

If x<3 then Ix-3I = 3-x and Ix-4I = 4-x, so Ix-3I + Ix-4I = 7-2x

Answer 6

curiously so. I actual have been violated for doing so. {shrug} specific, I actual have allegedly responded a question without particularly answering the question as consistent with the subjective interpretation of the Yahoo community rules.

Answer 7

1x-31-1x-41= 2x-72
if <3 then
with x=1, 2x-72= -70
with x=2, 2x-72=-68