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Modulus Function Question

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Don't understand the question ☹️
(Ignore bottom question).
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Avatar for Notnek
Notnek
21
Original post by TSRforum
image.jpeg
Don't understand the question ☹️
(Ignore bottom question).

A different example which will give you a hint:

x^2 is always positive so | x^2 | is identical to x^2 since taking the absolute value has no effect for a positive number.

Now what can you say about 2x - 1 when x > 0.5 ?
Reply 2
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TSRforum
OP
7
Original post by notnek
A different example which will give you a hint:

x^2 is always positive so | x^2 | is identical to x^2 since taking the absolute value has no effect for a positive number.

Now what can you say about 2x - 1 when x > 0.5 ?


Y is greater than zero?

Still don't understand ..


Posted from TSR Mobile
(edited 8 years ago)
Reply 3
Avatar for Notnek
Notnek
21
Original post by TSRforum
Y is greater than zero?

Still don't understand ..


Posted from TSR Mobile

When x > 0.5, 2x - 1 is always positive (check this).

So | 2x - 1 | is the same as 2x - 1.

S y = x + | 2x - 1 | is the same as _____________
Reply 4
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TSRforum
OP
7
Original post by notnek
When x > 0.5, 2x - 1 is always positive (check this).

So | 2x - 1 | is the same as 2x - 1.

S y = x + | 2x - 1 | is the same as _____________


3x-1?


Posted from TSR Mobile
Reply 5
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Notnek
21
Original post by TSRforum

Correct.
Reply 6
Avatar for B_9710
B_9710
18
When x>1/2, |2x-1|=2x-1
When x<1/2, |2x-1|=1-2x
Reply 7
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TSRforum
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7
Original post by B_9710
When x>1/2, |2x-1|=2x-1
When x<1/2, |2x-1|=1-2x


What about this question:

Find the solution of eq. |x-3| + |x+3| = 6

(Requires sketching y=|x|, y=|x-3| and y= |x-3| + |x+3|)


Posted from TSR Mobile
Reply 8
Avatar for B_9710
B_9710
18
Original post by TSRforum
What about this question:

Find the solution of eq. |x-3| + |x+3| = 6

(Requires sketching y=|x|, y=|x-3| and y= |x-3| + |x+3|)


Posted from TSR Mobile


Split it up. Consider what happens to the modulus functions when x>3, when x<-3, when -3<x<3.
Reply 9
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TSRforum
OP
7
Original post by B_9710
Split it up. Consider what happens to the modulus functions when x>3, when x<-3, when -3<x<3.

Don't understand what you mean...


Posted from TSR Mobile
Reply 10
Avatar for TSRforum
TSRforum
OP
7
Original post by TSRforum
Don't understand what you mean...


Posted from TSR Mobile


Also how would I go about sketching y= |x-3| + |x+3|?


Posted from TSR Mobile
Original post by TSRforum
image.jpeg
Don't understand the question ☹️
(Ignore bottom question).


|x|= x for x>=0 pr -x for x<0
When is 2x-1>=0 or <0?


Posted from TSR Mobile
Original post by TSRforum
Also how would I go about sketching y= |x-3| + |x+3|?


Posted from TSR Mobile


Split between regions x>3, x<-3 and -3<x<3


Posted from TSR Mobile
Reply 13
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TSRforum
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Original post by physicsmaths
|x|= x for x>=0 pr -x for x<0
When is 2x-1>=0 or <0?


Posted from TSR Mobile


Already done the question 🙂
Need help with the second question I posted


Posted from TSR Mobile
Reply 14
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TSRforum
OP
7
Original post by physicsmaths
Split between regions x>3, x<-3 and -3<x<3


Posted from TSR Mobile


Don't understand...


Posted from TSR Mobile
Reply 15
Avatar for B_9710
B_9710
18
Original post by TSRforum
Don't understand...


Posted from TSR Mobile


When x>3, |x-3|=x-3 and |x+3|=x+3.
Use this logic to show what happens to each of the modulus functions in the 3 regions I said earlier.
Make sense or still not sure?
Original post by B_9710
When x>3, |x-3|=x-3 and |x+3|=x+3.
Use this logic to show what happens to each of the modulus functions in the 3 regions I said earlier.
Make sense or still not sure?


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