Inequalities (1 Viewer)

InteGrand · Mar 25, 2016

Nailgun said:
$$Q. Solve $ \left | 1 - |x| \right | \geq 1$

$$Could someone help me with this? I understand the cases and stuff, and it works out to: $ x\ge 0,\quad x\le 0,\quad x\le -2,\quad x\ge 2$

$What I don't get is how you resolve all of these, like what is the rule/logic? I thought the first two would just imply that the domain was all real x, and so the second two inequalities make up the solution, but apparently another solution is $x=0$? Is it like the intersecting portion?$

$\noindent I think it's decently easy to do graphically too (probably easier than algebraic method), as long as you're careful not to miss out on a solution. Basically the graph comes out to be this sort of W shape. We will see that the solutions are: $x\leq -2$, $x=0$ (this is the one easy to miss), or $x\geq 2$.$

Nailgun · Mar 25, 2016

InteGrand said:
$\noindent I think it's decently easy to do graphically too (probably easier than algebraic method), as long as you're careful not to miss out on a solution. Basically the graph comes out to be this sort of W shape. We will see that the solutions are: $x\leq -2$, $x=0$ (this is the one easy to miss), or $x\geq 2$.$

Ah okay, so it's sort of like a piece-wise function with x<0 and x>0
and then you test solution for x=0?

Even then, in terms of the inequalities how does one do that algebraically?
I've had this issue before with a marathon question I think

edit: okay i dont think its like a piecewise function lol

InteGrand · Mar 25, 2016

Nailgun said:
$$Q. Solve $ \left | 1 - |x| \right | \geq 1$

$$Could someone help me with this? I understand the cases and stuff, and it works out to: $ x\ge 0,\quad x\le 0,\quad x\le -2,\quad x\ge 2$

$What I don't get is how you resolve all of these, like what is the rule/logic? I thought the first two would just imply that the domain was all real x, and so the second two inequalities make up the solution, but apparently another solution is $x=0$? Is it like the intersecting portion?$

$\noindent Yes, $x=0$ is a solution, since we get equality in that case, as seen by direct substitution into the L.H.S. Here is an algebraic method to solve this inequation.$

$\noindent \underline{\large Case 1. $x \geq 0\Longleftrightarrow |x| = x$}$

$\noindent Now the inequation is $|1-x| \geq 1$, $x \geq 0$ So we need two subcases.$

$\noindent \underline{\small Case 1a. $|1-x| = 1-x \Longleftrightarrow 1-x \geq 0\Longleftrightarrow x \leq 1$}$

$\noindent In Case 1a, we have $0 \leq x \leq 1$, and the inequation is $1-x \geq 1$. So $x \leq 0$. This occurs in the range $0\leq x \leq 1$ if and only if $x=0$. Hence one solution is \boxed{$x=0$}.$

$\noindent \underline{\small Case 1b. $|1-x| = x-1 \Longleftrightarrow 1-x \leq 0\Longleftrightarrow x \geq 1$}$

$\noindent In Case 1b, we have $x \geq 1$, and the inequation is $x-1 \geq 1$. This is equivalent to $x \geq 2$. This occurs in for any $x\geq 1$. Hence the solution set for Case 1b is \boxed{$\{x : x \geq 2\}$}.$

$\noindent Now we could go ahead and consider Case 2 ($x < 0$) and do similar things, but a shortcut is that the L.H.S. is a function that is clearly even, since if we replace $x$ with $-x$, it stays the same. So the function's graph is symmetric about the $y$-axis, and so the solution sets for the inequality are symmetric about the $y$-axis too. So by doing a similar analysis for Case 2, we'd get this additional solution set (just the solution in Case 1b flipped about the $y$-axis): \boxed{$\{x : x \leq -2\}$}.$

$\noindent To sum up, the solutions are $x=0$, $x \geq 2$ or $x\leq -2$.$

InteGrand · Mar 25, 2016

Nailgun said:
Ah okay, so it's sort of like a piece-wise function with x<0 and x>0
and then you test solution for x=0?

Even then, in terms of the inequalities how does one do that algebraically?
I've had this issue before with a marathon question I think

edit: okay i dont think its like a piecewise function lol

$\noindent If we sketch the graph correctly, we don't need to actually `test' the solution of $x=0$ (i.e. it will come straight from our sketch rather than having to do a separate test to `find' it), because it's clear that the `middle' of the `W' will be at $x=0,y=1$ (of course, it's good to check after finding $x=0$ as a solution that it indeed works by subbing it in to the L.H.S.). The only thing we need to do for the graphical methods is find where the line $y=1$ cuts the `outer' parts of the W shape. It is clear by basically testing $x=2$ that the points where these happen are at $x=\pm 2$.$

$\noindent The reason we'd think $2$ would be where this happens is that the line segments / rays forming the W shape all have slope $1$ (in absolute value), because they come from flipping and translating parts of the graph of $y=|x|$. Since it is clear that $(1,0)$ is a point on the W, and we can see it has slope 1 to the right of this, it is clear that at $x=2$, we will have $y$ increment up by one unit to 1.$$

$\noindent By the way, the way to sketch $y=f(x)=\big {|} 1 - |x|\right \big{|}$ is pretty systematic too, though it may involve some graphical transformations (mainly the absolute value transformation) that are more familiar to the average 4U student than the average 2U student (since there's a topic on Graphs in 4U). First, we sketch $y=|x|$ (standard graph). Then we get the graph of $y=-|x|$ by flipping that of $y=|x|$ about the $x$-axis. Then, we obtain the graph of $y=1-|x|$ by shifting the last graph up by one unit. Finally, we get the graph of $y=f(x)=\big {|} 1 - |x|\right \big{|}$ by flipping the parts of the previous graph that were below the $x$-axis about the $x$-axis (and leaving other parts unchanged).$

Nailgun · Mar 25, 2016

InteGrand said:
$\noindent \underline{\large Case 1. $x \geq 0\Longleftrightarrow |x| = x$}$

Sorry, could you explain this line a bit further? I understand the rest, but I don't get how we make the statement |x|=x
what if x=-2

Drongoski · Mar 25, 2016

InteGrand said:
$\noindent I think it's decently easy to do graphically too (probably easier than algebraic method), as long as you're careful not to miss out on a solution. Basically the graph comes out to be this sort of W shape. We will see that the solutions are: $x\leq -2$, $x=0$ (this is the one easy to miss), or $x\geq 2$.$

Indeed - I just missed the isolated x = 0!! I overlooked this case when I looked at the 2 sets of solutions I'd sketched on the number line.

Nailgun · Mar 25, 2016

InteGrand said:
$\noindent By the way, the way to sketch $y=f(x)=\big {|} 1 - |x|\right \big{|}$ is pretty systematic too, thought it may involve some graphical transformations (mainly the absolute value transformation) that are more familiar to the average 4U student than the average 2U student (since there's a topic on Graphs in 4U). First, we sketch $y=|x|$ (standard graph). Then we get the graph of $y=-|x|$ by flipping that of $y=|x|$ about the $x$-axis. Then, we obtain the graph of $y=1-|x|$ by shifting the last graph up by one unit. Finally, we get the graph of $y=f(x)=\big {|} 1 - |x|\right \big{|}$ by flipping the parts of the previous graph that were below the $x$-axis about the $x$-axis (and leaving other parts unchanged).$

Oh right thanks, that makes a lot of sense.

That was basically what I was trying to do to graph it, except I didn't recognize that you have to flip the parts below the x-axis

InteGrand · Mar 25, 2016

Nailgun said:
Sorry, could you explain this line a bit further? I understand the rest, but I don't get how we make the statement |x|=x
what if x=-2

$\noindent This is when $x\geq 0$ as the Case 1 line says, so $x=-2$ isn't under consideration. In other words, we're considering the case where $|x|$ is just the same as $x$, and we're saying this is equivalent to $x$ being non-negative (which is true, since the definition of the absolute value function is:$

$$|\xi|=\begin{cases} \xi, & \text{ if } \xi \geq 0 \\ -\xi, & \text{ if } \xi < 0 \, (\text{or we can have } \leq \text{here, makes no difference}) \end{cases}$).$

$\noindent So usually when we have equations or inequations involving absolute values, the cases we want to consider are (1) when the absolute value is equal to its argument ($|\xi| = \xi \Longleftrightarrow \xi \geq 0$), and (2) when the absolute value is equal to negative its argument ($|\xi| = -\xi \Longleftrightarrow \xi \leq 0$). Subcases can arise when we have multiple absolute value signs present.$

Drongoski · Mar 25, 2016

Another way to look at it:

2 cases:

i) x >= 0

inequation is simply: |1 - x| >= 1 which yields: x <= 0 or x >= 2

ii) x < 0

inequation becomes: |1 + x| >= 1 which yields: x <= -2 and x >= 0

Taking the intersection of these 2 sets: we get x <= -2. x = 0 and x >= 2

Here intersection means all the values of x which satisfy (or are common to) the above 2 cases.

Nailgun · Mar 25, 2016

InteGrand said:
$\noindent This is when $x\geq 0$ as the Case 1 line says, so $x=-2$ isn't under consideration. In other words, we're considering the case where $|x|$ is just the same as $x$, and we're saying this is equivalent to $x$ being non-negative (which is true, since the definition of the absolute value function is:$

$$|\xi|=\begin{cases} \xi, & \text{ if } \xi \geq 0 \\ -\xi, & \text{ if } \xi < 0 \, (\text{or we can have } \leq \text{here, makes no difference}) \end{cases}$).$

$\noindent So usually when we have equations or inequations involving absolute values, the cases we want to consider are (1) when the absolute value is equal to its argument ($|\xi| = \xi \Longleftrightarrow \xi \geq 0$), and (2) when the absolute value is equal to negative its argument ($|\xi| = -\xi \Longleftrightarrow \xi \leq 0$). Subcases can arise when we have multiple absolute value signs present.$

Okay, right. I've got it now I'm pretty sure. I used case 1 and 2 as -(1-|x|) and 1-|x| and the |x| as the subcases, but I didn't consider that saying |1-x|=1-x is also saying 1-x>0
Alright, thank you

InteGrand · Mar 25, 2016

Drongoski said:
Another way to look at it:

2 cases:

i) x >= 0

inequation is simply: |1 - x| >= 1 which yields: x <= 0 or x >= 2

ii) x < 0

inequation becomes: |1 + x| >= 1 which yields: x <= -2 and x >= 0

Taking the intersection of these 2 sets: we get x <= -2. x = 0 and x >= 2

Here intersection means all the values of x which satisfy (or are common to) the above 2 cases.

Also, when we get the x <= 0 case in case i), we can immediately say this gives us just x = 0, since case i) had x >= 0. In other words, for each case, we take the intersection of the solution sets with the condition for the case (e.g. for case i), we take the intersection of ''x <= 0 or x >= 2'' with the condition ''x >= 0'', which leaves us with ''x = 0 or x >= 2''). Then we do this for case ii) as well, and finally take the union of these. (Here, we have meant 'union' and 'intersection' in the usual sense.)

Nailgun · Mar 25, 2016

Drongoski said:
Another way to look at it:

2 cases:

i) x >= 0

inequation is simply: |1 - x| >= 1 which yields: x <= 0 or x >= 2

ii) x < 0

inequation becomes: |1 + x| >= 1 which yields: x <= -2 and x >= 0

Taking the intersection of these 2 sets: we get x <= -2. x = 0 and x >= 2

Here intersection means all the values of x which satisfy (or are common to) the above 2 cases.

Yeah this is what I did initially, but I didn't realize that you take the intersection of the x <= 0, x >= 0
Much appreciated both of you

tbh inequalities seem so counter intuitive lel, i can visualize the number-line so it makes sense but at the same time saying x is both greater than 2 but also less than -2 seems so counter intuitive. i think that's where I get confused

InteGrand · Mar 25, 2016

Nailgun said:
Yeah this is what I did initially, but I didn't realize that you take the intersection of the x <= 0, x >= 0
Much appreciated both of you

tbh inequalities seem so counter intuitive lel, i can visualize the number-line so it makes sense but at the same time saying x is both greater than 2 but also less than -2 seems so counter intuitive. i think that's where I get confused

I think this question was from a Carrotsticks Trial, designed to trip people up. If it came up in the 2U HSC paper, it'd probably have a few sub-parts to it.

Nailgun · Mar 25, 2016

InteGrand said:
I think this question was from a Carrotsticks Trial, designed to trip people up. If it came up in the 2U HSC paper, it'd probably have a few sub-parts to it.

It was a question he scrapped for being too easy

That's true, but I still feel like I should've been able to do it ahahah

Drongoski · Mar 25, 2016

You'll never get such a question in 2U Maths - it'd be too hard. For 3U, perhaps.

InteGrand · Mar 25, 2016

Yeah. For most questions in nU Carrotsticks trials, they would probably only appear in (n+1)U (or higher) papers

Nailgun · Mar 25, 2016

InteGrand said:
Yeah. For most questions in nU Carrotsticks trials, they would probably only appear in (n+1)U (or higher) papers

So where do they test questions for the 4U paper lol

InteGrand · Mar 25, 2016

Nailgun said:
So where do they test questions for the 4U paper lol

In the mythical land of 5U

Paradoxica · Mar 25, 2016

Lol... That one time the discussion about vehicles happened...

I thought MX3 was Extension 3...

Inequalities (1 Viewer)

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