Question

find the values of b and c which will make f(x) continuous for all real numbers if
f(x) = x+1 abs(x-2) <1
x^2+bx+c abs(x-2) >=1

Answer 1

its the absolute values that are throwing me off.
it's in the left braces
and the second one is greater than or equal to, sorry for any confusion.

Answer 2

first off what does
\[|x-2|<1\] really mean? it means
\[-1<x-2<1\] i.e.
\[1<x<3\] so we are only concerned with the points x = 1 and x = 3 where the definition of the function changes

Answer 3

so here is what you do. replace x by 1 in both formulas and set them equal. you get
\[1+1=1+b+c\]

Answer 4

so you know
\[b+c=1\]

Answer 5

To be continuous, the two pieces must equal the same value when one piece switches to another (i.e. - x={1,3}). Thus:\[(1)+1=(1)^2+b(1)+c \text{ and }(3)+1=(3)^2+b(3)+c\]\[1=b+c\text{ and } -5=3b+c\]The solution of this system of equations is:\[b=-3 \text{ and } c=4\]

Answer 6

now replace x by 3 in both formulas. you get
\[3+1=9+3a+c\] so
\[3a+c=-5\]

Answer 7

damn i used a instead of b sorry i will rewrite it

Answer 8

It's done already, dude.

Answer 9

that makes so much sense, thank you!

Answer 10

you get
\[3+1=9+3b+c\] so
\[3b+c=-5\] now solve the system
\[b+c=1\]
\[3b+c=-5\] etc