Question

sketch a graph for the function;
f(x)=1/2(x-|x|)

Answer 1

For x > 0, |x| = x and x - |x| = x - x = 0
What about x < 0?

Answer 2

hmm....I am not sure

Answer 3

You'll have to do better than that.

For x < 0, |x| = -x

Do you believe this?

Answer 4

lt's the opposite when it is less

Answer 5

|-3| = -(-3) = +3

|-6| = -(-6) = +6

You must be certain that this is so. For x < 0, |x| = -x.

Answer 6

(x-|x|)
the x-value must be less than zero. That means when you're sketching the curve, you now know that the positive side of the x-axis cannot be used in this instance.

Answer 7

?? The Domain is \(\mathbb{R}\). There is no restriction.

For \(x \ge 0\), |x| = x and we have x - |x| = x - (x) = 0
For \(x < 0\), |x| = -x and we have x - |x| = x - (-x) = x + x = 2x

Answer 8

No, I don't think so.I'm sure of it that if x<0, you would have a negative denominator.

Answer 9

For x<0 , (-x) - |(-x)|=-x-(x)=-x-x=-2x

Answer 10

The only denominator is 2.

@malibugranprix2000, do you mean \(\dfrac{1}{2}(x - |x|)\) as you have written, or did you intend \(\dfrac{1}{2(x-|x|)}\), which is not what you have written?

Answer 11

No, x is still x, no matter the sign of x.

Answer 12

No there's no absolute value sign on the x that's free.

Answer 13

If x was negative. you would have -2x regardless of anything weird happening.

Answer 14

No, if x is negative, then 2x would be a negative number. Why would it suddenly change to positive (-2x)?

Answer 15

What are you on about? What you said was practically the same. negative 2x is the same as positive (-2x)....-2x=+(-2x)?

Answer 16

@malibugranprix2000 Please draw your correct equation.

Answer 17

No, that is just wrong.

For x < 0, 2x < 0 and -2x > 0

Answer 18

Ah okay. Yep, you're right. Sorry about that mate.

Answer 19

It is 1/2(x-|x|) not 1/2....

Answer 20

it is not 1 over 2(x-|x|)

Answer 21

Okay. I think tkhunny's got it here then.

Answer 22

You wrote the same thing, which is \(\dfrac{1}{2}(x - |x|)\). Please refer to this as #1.

If you mean \(\dfrac{1}{2(x-|x|)}\), please consider the Order of Operations and use more parentheses. Please refer to this as #2.

Which is it?

Many thanks to @Azteck for pointing out the possible ambiguity.

Answer 23

number 1

Answer 24

For #1, we have:

For x≥0 , |x| = x and we have x - |x| = x - (x) = 0 an horizontal line on the non-negative x-axis.
For x<0 , |x| = -x and we have x - |x| = x - (-x) = x + x = 2x a line descending from the Origin.

For #2, we have

For x≥0 , |x| = x and we have x - |x| = x - (x) = 0 and this is not in the Domain.
For x<0 , |x| = -x and we have x - |x| = x - (-x) = x + x = 2x and we have half an hyperbola.

An interesting problem, either way.

Answer 25

thanks @tkhunny and @Azteck for the explanation