Question

Let f(x) = lx-1l. Which of the following conditional statements hold true?

*Check all that apply*

A. f(x) = x - 1 if x > 1
B. f(x) < 0 if x = 0
C. f(x) = x - 1 if x > 0
D. f(x) = -(x - 1) if x < 1

Answer 1

do you know the definition of \(|x-1|\) ?

Answer 2

i mean the real math definition, not the "make it positive" definition

Answer 3

No. But an explanation would be much appreciated :D

Answer 4

ok lets start with the real definition of \(f(x)=|x|\) which is
\[f(x) = |x| = \left\{\begin{array}{rcc}
-x & \text{if} & x <0 \\
x& \text{if} & x \geq 0
\end{array}
\right.
\]

Answer 5

a piecewise function reads
\(|x|\) is \(-x\) if \(x\) is negative (which of course makes \(-x\) positive) or just \(x\) if \(x\) is positive (since it is already positive, don't change it)

Answer 6

now we replace \(x\) by \(x-1\) to get your example
\[f(x) = |x-4| = \left\{\begin{array}{rcc}
-(x-1) & \text{if} & x-1 <0 \\
x-1& \text{if} & x-1 \geq 0
\end{array}
\right.
\]

Answer 7

we really want to write this as
\[f(x) = |x-4| = \left\{\begin{array}{rcc}
-(x-1) & \text{if} & x <1 \\
x-1& \text{if} & x \geq 1
\end{array}
\right.\]

Answer 8

in other words the absolute value of \(x-1\) is just \(x-1\) so long as \(x\geq 1\) i.e. as long as \(x-1\) is positive, but if \(x-1\) is negative then \(|x-1|\) is \(-(x-1)\)

Answer 9

now we can check all that apply
\(A. f(x) = x - 1 \) if \( x > 1\) YES for sure that is the second line of the definition

Answer 10

B is just plain silly, the absolute value is never negative, so NO

Answer 11

how about C?

Answer 12

I think C applies

Answer 13

Because x is greater then 0

Answer 14

?