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core 3 functions

f(x)=2-(x+1)^1/3
find the set of values of x for which f(x)=modulus f(x).
(2 marks)
Original post by Chelsea12345
f(x)=2-(x+1)^1/3
find the set of values of x for which f(x)=modulus f(x).
(2 marks)


Could you show how you approached the problem and at which step you are stuck?
Original post by Geodesic
Could you show how you approached the problem and at which step you are stuck?


so far i know that if it's a modulus, then the number inside can be positive or negative. But, im unsure how to solve this since the equation is equal to modulus function?
Original post by Chelsea12345
so far i know that if it's a modulus, then the number inside can be positive or negative. But, im unsure how to solve this since the equation is equal to modulus function?


Alright well so if we have x2−8=∣x2−8∣x^2 - 8 = |x^2 - 8| the two possible equations we could have would be x2−8=x2−8x^2 - 8 = x^2 - 8 andx2−8=−x2+8x^2 - 8 = -x^2 + 8.

So in the case of 2−(x+1)1/3=∣2−(x+1)1/3∣2 - (x+1)^{1/3} = |2 - (x+1)^{1/3}| , what are the two possible equations we can have to solve for xx ?
Original post by Geodesic
Alright well so if we have x2−8=∣x2−8∣x^2 - 8 = |x^2 - 8| the two possible equations we could have would be x2−8=x2−8x^2 - 8 = x^2 - 8 andx2−8=−x2+8x^2 - 8 = -x^2 + 8.

So in the case of 2−(x+1)1/3=∣2−(x+1)1/3∣2 - (x+1)^{1/3} = |2 - (x+1)^{1/3}| , what are the two possible equations we can have to solve for xx ?


it would be:
2-(x+1)^1/3= 2-(x+1)^1/3
and-(2-(x+1)^1/3 = 2-(x+1)^1/3
how do i solve after this? is it just making them equal to zero for both and solving or do i have to do something different?
Original post by Chelsea12345
it would be:
2-(x+1)^1/3= 2-(x+1)^1/3
and-(2-(x+1)^1/3 = 2-(x+1)^1/3
how do i solve after this? is it just making them equal to zero for both and solving or do i have to do something different?


Yes well for the 1st one, it is true for all x x .

For the second one, I would expand the brackets then simply just solve for x x as you would normally.
Original post by Geodesic
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Original post by Chelsea12345
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I think this thread's going off on the wrong track.

The modulus of something is equal to that something when that something is positive or zero.

Viz. y=|y| when y >=0

Now replace y with f(x).

f(x) equals its modulus when f(x) >= 0. So the question becomes when is f(X) >= 0
Whenever f(x) is positive (or zero), f(x) will be equal to mod(f(x)). On the other hand, when f(x) is negative....

Update: similar thought to @Ghostwalker.
(edited 6 years ago)
Original post by ghostwalker
I think this thread's going off on the wrong track.

The modulus of something is equal to that something when that something is positive or zero.

Viz. y=|y| when y >=0

Now replace y with f(x).

f(x) equals its modulus when f(x) >= 0. So the question becomes when is f(X) >= 0


That was exactly what I was trying to explain, but I approached it in slightly different way, which perhaps was wrong and longwinded. Sorry about that.

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