A maths - Modulus Functions
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Came across this question just now. Had a hard time trying to solve it. Do take a look and see where you are able to help.
Sketch the graph of y - 3 = -|3 - 2x| for -1< or equal to x < or equal to 4. Hence state the values of x for which y>2
Can't you multiply the - sign into |3-2x| and get |2x-3|? I understand that this would result in a different answer towards the end.
But how do you explain this if you were to use this method ( A correct way) to solve this question?
If f(x) = |x-2| + |4-2x|, show that f(x) = k|x-2|, where k is a constant to be determined.
so you do something like this : |x-2| + |-2||x-2| = 3|x-2|
Going back to my earlier question, why is it that you can't mutiply the -1 inside the term but you can factorise it out?
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thank you for posting..
me looking for ans toward modulus qns too >.<
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Originally posted by SBS7484P:
thank you for posting..
me looking for ans toward modulus qns too >.<
Haha. Modulus functions should be an easy topic by right. But I somehow have difficulties for no reason. >.< Argh, and do you have any idea how you can correctly and accurately dervie the corresponding values of y on the modulus graphs? I get that wrong sometimes. -
those requiring some in depth calculation to get the graph i dunno.
but those already pre-drawn, then they just ask u to "modulus it" i know.
everything below the zero mark on the y-axis, mirror it upwards.
damn. the last sentence wasent of much help eh.
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Originally posted by bonkysleuth:
Came across this question just now. Had a hard time trying to solve it. Do take a look and see where you are able to help.
Sketch the graph of y - 3 = -|3 - 2x| for -1< or equal to x < or equal to 4. Hence state the values of x for which y>2
Can't you multiply the - sign into |3-2x| and get |2x-3|? I understand that this would result in a different answer towards the end.
But how do you explain this if you were to use this method ( A correct way) to solve this question?
If f(x) = |x-2| + |4-2x|, show that f(x) = k|x-2|, where k is a constant to be determined.
so you do something like this : |x-2| + |-2||x-2| = 3|x-2|
Going back to my earlier question, why is it that you can't mutiply the -1 inside the term but you can factorise it out?
Because |3-2x| = |2x-3|, but both are not equal to -|3-2x| or -|2x-3|.
|4-2x| = |-2||x-2| = 2|x-2| (because |-2| = 2). Hence that statement is possible.
Whatever is already inside the modulus you can make it negative and it'll give you the same positive result, but making what is outside negative will not satisfy the equation anymore.
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Originally posted by SBS7484P:
those requiring some in depth calculation to get the graph i dunno.
but those already pre-drawn, then they just ask u to "modulus it" i know.
everything below the zero mark on the y-axis, mirror it upwards.
damn. the last sentence wasent of much help eh.
Ok. Lets try this. The question is : sketch the graph of y = |2x-1| and find the set of values of x for which y<3.
I got the correct graph. However when you do the workings to get the set of values of x for which y < 3, it goes as follows:
2x - 1 < 3
2x < 4
x<2
2x - 1 > -3
2x>-2
x>-1
Yea, i don't know how they got the highlighted part. why is 2x-1 more than -3? If you know, do explain to me. Otherwise, hopefully, other forumers will help.
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for e.g. |x| < 2 automatically translates to -2<x<2, which translates to x>-2 AND x<2.
so, for |2x-1| < 3, it must be -3<|2x-1|<3, which translates to 2x-1>-3 AND 2x-1<3.
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Originally posted by ^tamago^:
for e.g. |x| < 2 automatically translates to -2<x<2, which translates to x>-2 AND x<2.
so, for |2x-1| < 3, it must be -3<|2x-1|<3, which translates to 2x-1>-3 AND 2x-1<3.
That's EXACTLY the part I don't get. if |x| < 2... like those normal questions you do, shouldnt it be x<2 or x<-2? -
smaller than -2 means all the way till minus infinity leh!
modulus/absolute/numerical value less than 2 (|x| < 2) consist of numbers like -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5 and 2.
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Originally posted by ^tamago^:
smaller than -2 means all the way till minus infinity leh!
modulus/absolute/numerical value less than 2 (|x| < 2) consist of numbers like -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5 and 2.
O, think I sort of get the idea now. Though I'm still a little perplexed by it all. Hey, can you help draw or describe the graph that is supposed to be drawn for the above the question? I mean, the first question I asked. The one with the long stretch of equations. Cuz we are supposed to draw THREE graphs in just ONE question. Kind of funny. -
total confusion.
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Originally posted by SBS7484P:
total confusion.
will try to simplify the explanation (if possible) later on. But for now, I have another question on Modulus Functions. lol.
Sketch, on the same diagram, the graphs of y = 4|x| and y = |x-1|. state the number of solutions of the equation 4|x| = |x-1| and hence find the solution if 4|x| > |x-1|
The answer sheet writes:
-4x = -x + 1
-3x = 1
x = - 1/3
-4x = x-1
-5x = -1
x = 1/5
Goes on to state that x < - 1/3 or x >1/5
Just wondering why it knows that x is smaller than -1/3 and subsequently x > 1/5 when it has been using the " = " sign all the while...strange.
why can't it use this method : (leave out the minus sign before 4x)
4x > x -1
3x > -1
x > -1/3
4x < 1-x
5x <1
x<1/5
the value's the same. only exception is the < or > sign. but is there any error in the usage of signs in my working? if there is, i think i might have some misconception of some ideas here.
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Red is what is in the question, blue is if you add a negative sign outside the modulus. -
Originally posted by bonkysleuth:
will try to simplify the explanation (if possible) later on. But for now, I have another question on Modulus Functions. lol.
Sketch, on the same diagram, the graphs of y = 4|x| and y = |x-1|. state the number of solutions of the equation 4|x| = |x-1| and hence find the solution if 4|x| > |x-1|
The answer sheet writes:
-4x = -x + 1
-3x = 1
x = - 1/3
-4x = x-1
-5x = -1
x = 1/5
Goes on to state that x < - 1/3 or x >1/5
why can't it use this method : (leave out the minus sign before 4x)
4x > x -1
3x > -1
x > -1/3
4x < 1-x
5x <1
x<1/5
Surely you can, but it is a case of "Hence, or otherwise..." as seen in many questions.
The objective of making you sketch both of them on the same graph is so that you can compare directly at which points is the graph of f(x)=4|x| at a higher value of y than f(x)=|x-1|. Surely there won't be lost marks if you decided to solve the second part of the question directly, but since it is easier to err compared to looking at the graph, if you end up making a mistake, there goes your marks. -
Originally posted by ^tamago^:
Surely you can, but it is a case of "Hence, or otherwise..." as seen in many questions.
The objective of making you sketch both of them on the same graph is so that you can compare directly at which points is the graph of f(x)=4|x| at a higher value of y than f(x)=|x-1|. Surely there won't be lost marks if you decided to solve the second part of the question directly, but since it is easier to err compared to looking at the graph, if you end up making a mistake, there goes your marks.
Thanks so much! Where do you get the software for drawing that graph by the way? -
Mac OS X's Grapher.