Practicing Success
If $m$ and $M$ are respectively minimum and maximum values of $f(x)= |2-|x||, -3≤x≤3$, then: |
m = 0 and M = 2 m = 1 and M = 2 m = 0 and M = 4 m = 0 and M = 1 |
m = 0 and M = 2 |
$f(x)= |2-|x||,\,\, -3≤x≤3$ $⇒0≤|x|≤3$ $⇒-3≤-|x|≤0$ $⇒-1≤2-|x|≤2$ so $0≤|2-|x||≤2$ m = 0 and M = 2 |