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    If you remove the absolute value sign from x, from what i learnt so far, you get two values for x i.e +ve and -ve x.
    So my intuition tells me there should be 4 combinations which are
    1) (x-1)(x-2) gives a minimun with critical values of x=1 and x=2 , y- intercept at y=2

    2) (-x-1)(-x-2) also gives a minimun which is basically the reflection of the graph above in the y-axis.
    So this is what the answer in the textbook provided but i don't agree with it.

    Isnt there should be 2 more combinations? I.e
    (-x-1)(x-2) and (x-1)(-x+2)
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    Well if you define f(x) = (x-1)(x-2) then f(|x|) = (|x|-1)(|x|-2). This should help you.
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    (Original post by Bilbao)
    If you remove the absolute value sign from x, from what i learnt so far, you get two values for x i.e +ve and -ve x.
    So my intuition tells me there should be 4 combinations which are
    1) (x-1)(x-2) gives a minimun with critical values of x=1 and x=2 , y- intercept at y=2

    2) (-x-1)(-x-2) also gives a minimun which is basically the reflection of the graph above in the y-axis.
    So this is what the answer in the textbook provided but i don't agree with it.

    Isnt there should be 2 more combinations? I.e
    (-x-1)(x-2) and (x-1)(-x+2)
    The way i think about it is knowing what the transformation of f(|x|) does. If you know y = f(x) then y = f(|x|) has a specific transformation of the first graph, I'd search up because i can't rly explain it well on here.

    Then once you understand what that does just think about f(x) = (x-1)(x-2), which you should probably know how to sketch, and how then what f(|x|) must look like.
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    (Original post by Bilbao)
    If you remove the absolute value sign from x, from what i learnt so far, you get two values for x i.e +ve and -ve x.
    So my intuition tells me there should be 4 combinations which are
    1) (x-1)(x-2) gives a minimun with critical values of x=1 and x=2 , y- intercept at y=2

    2) (-x-1)(-x-2) also gives a minimun which is basically the reflection of the graph above in the y-axis.
    So this is what the answer in the textbook provided but i don't agree with it.

    Isnt there should be 2 more combinations? I.e
    (-x-1)(x-2) and (x-1)(-x+2)
    There are only two cases, where x>=0 in which case |x|=x and x<0 in which case |x|=-x.
    How would you get (x-1)(-x+2) as you're saying, you're saying that in on every bracket |x|=x which means that x>=0 and in the other you're saying |x|=-x meaning x<0 so basically you have that x>=0 and x<0 which doesn't make sense.
    Hope this helps.
 
 
 

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