Absolute Value

[smsuski]

1. lx-al=b where a and b are positive
this is what i have for the anwsers so far

-x+a=b x-a=b
-x=b-a x=b+a
(-1)-x=(-1)b-a) x=a+b
x=a-b

2.lx+al(</=)a where a is positive

x+a(</=)a -x-a(>/=)a
x(</=)a-a -x(>/=)a+a
x(</=) 0 x(</=)-a-a
x(</=)-2a

3. Use Absolute Values to write the following statement more compactly: Whenever x is within .02 of 7, f(x) differs from 19 by no more than .3
the with in part has me confused. Here is what I have come up with so far.

for 7 (6.98</=x</=7.02) for 19 (18.97</=19.03)

F(x)= |(6.98</=x</=7.02)-(18.97</=19.03)|

 

[pka]

lx-al=b where a and b are positive
Then (x−a)=b or (x−a)=−b.
Thus x=a±b

|x+al≤a where a is positive
then −a≤(x+a)≤a
so −2a≤x≤0.

3. Use Absolute Values to write the following statement more compactly: Whenever x is within .02 of 7, f(x) differs from 19 by no more than .3
|x−7|≤0.2 implies |f(x)−19|≤0.3.