# Thread: Related range of values of f(x) necessary/sufficient/ with absolute values

1. ## Related range of values of f(x) necessary/sufficient/ with absolute values

Hi Forum.

But here I go again:

Consider two conditions
$x^2-3x-10$ and $|x-2|

Where x is a real number, and a is a positive real number.

The range of a, so that $|x-2| is a necessary condition for $x^2-3x-10<0$

The range of a, so that $|x-2| is a sufficient condition for $x^2-3x-10<0$

This problem is mainly about logic, right?
Here is something that I thought, after calculating the range of $x^2-3x-10<0$, that is $-2

If we have $-2
-2________________________________5
|________________________________|
___________________________a=0 ______________________________ a=3
|_________________________________________________ _| <-|x-2|<a

Edit: This didn't work so well but, the a=0 is right below -2 and the a=3 is right below 5

This is our sufficient condition: 0<a<3
I'm not sure what the absolute value does here.
What does the absolute value mean?
Absolute values are only the distance to x=0, right?

After calculating -correctly- you get a nicer solution that is:
$0

But how do we get such solution?

$|x-2| is
$x-2$ if $x-2\geqslant a$ and
$-x+2$ if $x-2

This relates to two range of values so this got a little out of control.
Somehow confusing, I wish I knew somewhere to find exercises of this kind!

I'm confused about that necessary/sufficient part

Sufficient
If you mow the lawn, you receive 10 dollars.
____________________Necessary

The $0 range is sufficient for f(x) to work, or something like that?
But where does necessity comes in this? I know that if something is necessary it is not sufficient.

A--->B
A implies B
A sufficient, B necessary

I appreciate any help!
Thanks!

2. Interesting problem. Your problem is dealing with implication. Let's let $P$ be the proposition that

$x^{2}-3x-10<0,$

and let's let $Q(a)$ be the proposition that

$|x-2|

Now, if you want to find the range of $a$ such that $Q(a)$ is sufficient for $P,$ then that's the same thing as finding the range of $a$ such that $Q(a)$ implies $P.$

Conversely (literally!), if you want to find the range of $a$ such that $Q(a)$ is necessary for $P,$ then that's the same thing as finding the range of $a$ such that $P$ implies $Q(a).$ So here, you just reverse the direction of the implication.

I agree with your 'sufficient' answer. $0 is exactly correct.

But now, with the 'necessary' answer, you want $P$ to imply $Q(a).$ How could that happen?