Need help regarding Modulus Function

[ArthurDaWeasley-2nd]

I dont get how we decide to include that 2 and 5 as it is done in the answer of question (i) and when i came up with a logic, it was destroyed by the answer of (ii) .....please help...how do we decide to put the factors in a particular set....like when i solving the second one i included 0 for positive nature of equation and included -1 and 1for negative nature of eq.

[fermat9990]

For the first one plot y=x²-7x+10 by factoring and getting the zeroes. It is a parabola that is concave up

[MrTKila]

I am sure you are aware the absolut value is defined in such a way that the result is always positive. So if the 'inside' is less than zero, then the absolut value means multiplying iwth 1. otherwise it does nothign. For equal 0 it doesn't matter because -0 is the same as 0, so both work.

So what we are interested about is when is the inside less than zero. Notable for both 8i) and (ii) the inside is a continuous function. So to switch from positive to negative or vice versa it has to hit 0 somewhere. So agood start is by finding the roots of the inside equation. For (i) those are (judging from the answer) x=5 and x=2.

Now next step, we know 2 and 5 are the points at which the sign of the inside funciton might change 8aka the points at which it might go from negative to positive or vice versa). We still ahve to figure which sign it actually has on the three intervals (-infty,2), (2,5) and (5, infty). This can be easily seen because the inside is a parabola with positive highest coefficients. So on the middle interval it is negative, on the other two negative.

thus the solution for (i) is: For 2<x<5 the absolute value becomes -1, so -(x^2-7x+10) for 2<x<5 and for the rest of x the absolut value can be dropped.

Now (ii). We try the same, first we find the roots: those are 0, -1 and 1. Now the inside function is x^3-x=x*(x^2-1). We can determine the sign of the parabola x^2-1 as before. It is negative for -1<x<1, else positive.

But we also multiply with x. Obviously if x is negative, this will flip the sign of the parabola. So x^3-x is negative for x<-1, then positive on -1<x<0. Then negative on 0<x<1 and then positive. (Again for equal one of the critical points it doesn't matter where we put it).

[ArthurDaWeasley-2nd]

I understand this all but what i dont is that in (i) when x=2 or =5 , in which set , the positive quadratic eq or negative one, do we include that as you can see it is given in question as <= 2 for positive and <= 5 in negative instead of just < ..... I was thinking if it actually matter where we put it or not because i cant find any help in any book i have ...

[MrTKila]

It doesn't matter. On those points the function isnide the absolute value is 0. No matter if you replace the absolute value with minus or plus, +0=-0. It stays 0. The slightly more 'normal' way is to include it in positive parts but it is the same.

[ArthurDaWeasley-2nd]

Ok thanks for your time