# 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
$\displaystyle x^2-3x-10$ and $\displaystyle |x-2|<a$

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

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

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

This problem is mainly about logic, right?
Here is something that I thought, after calculating the range of $\displaystyle x^2-3x-10<0$, that is $\displaystyle -2<x<5$

If we have $\displaystyle -2<x<5$
-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:
$\displaystyle 0<a\leqslant 3$

But how do we get such solution?

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

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 $\displaystyle 0<a\leqslant 3$ 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 $\displaystyle P$ be the proposition that

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

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

$\displaystyle |x-2|<a.$

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

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

I agree with your 'sufficient' answer. $\displaystyle 0<a\le 3$ is exactly correct.

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