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Quick Help with C3 Modulus inequality please. watch

1. So first asked to sketch y=[x] and y=[2x-4] which I have done so. Then asked to solve the inequality to which I got x=4 and x=4/3. Its the next bit I don't quite understand it says; "Hence, or otherwise, solve the inequality [x]>[2x-4]". and Im not even sure how to use my previous solutions for this because isnt the graph os 2x-4 greater than x? :/ Any help is greatly appreciated thanks.
2. Are you sure you were not meant to find the points of intersection?
3. Intersection (4,4), and below that intersection x is grater than 2x-4. Solution:
X < 4
4. (Original post by moz4rt)
Are you sure you were not meant to find the points of intersection?
No this is exactly what the questions says:
5. (Original post by 214508)
Intersection (4,4), and below that intersection x is grater than 2x-4. Solution:
X < 4
But don't they intersect at 2 points? at x=4/3 too...
6. Could you show me your graphs so i can see what youve done?

Edit: You should know that the modulus function takes whats on the inside and makes it positive
7. (Original post by MrToodles4)
So first asked to sketch y=[x] and y=[2x-4] which I have done so. Then asked to solve the inequality to which I got x=4 and x=4/3. Its the next bit I don't quite understand it says; "Hence, or otherwise, solve the inequality [x]>[2x-4]". and Im not even sure how to use my previous solutions for this because isnt the graph os 2x-4 greater than x? :/ Any help is greatly appreciated thanks.
If you've solved the equation, you have the critical points of the inequality. Should be pretty straightforward to get the regions where the inequality is true.

Also, note that it says "or otherwise", so if all else fails, you can solve it from scratch by squaring both sides (anything real squared is positive) and removing the modulus symbols.
8. (Original post by TheMindGarage)
If you've solved the equation, you have the critical points of the inequality. Should be pretty straightforward to get the regions where the inequality is true.

Also, note that it says "or otherwise", so if all else fails, you can solve it from scratch by squaring both sides (anything real squared is positive) and removing the modulus symbols.
I got 4/3<x<4 is this alright?
9. (Original post by MrToodles4)
I got 4/3<x<4 is this alright?
Looks right to me. In the exam, you can always check by substituting your critical values into the inequality, plus values in between to check that you have the right region (in this case, substitute x=4/3 and x=4 to see if both sides are equal, plus something convenient that's in between such as x=2 to check that the inequality is satisfied).
10. (Original post by TheMindGarage)
Looks right to me. In the exam, you can always check by substituting your critical values into the inequality, plus values in between to check that you have the right region (in this case, substitute x=4/3 and x=4 to see if both sides are equal, plus something convenient that's in between such as x=2 to check that the inequality is satisfied).
11. Always sketch the graphs with modulus first and check which part of the graph u want- i find it easier this way
12. you need to see where the graph of |x| is higher than the graph of |2x - 4|
13. (Original post by N/A IS THE NAME)
Always sketch the graphs with modulus first and check which part of the graph u want- i find it easier this way
Yeah I did sketch the graph as the questions earlier required it but thanks

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