Calculus help!!! Please help me!!!!!?

Question

i don't get this question at all, i don't know if it has to do with the limits or something. please help me!!!
here's the question=)
Determine the values of b & c so that the following function is continuous on the entire real line
f(x)= x + 1 1 < x < 3
x^2 + bx+ c |x - 2| >= 1

please help me!!

Answer 1

To solve this, we need the following condition to be true:

f(1) =

lim f(x)
x -> 1-
{i.e. as x approaches 1 from the left, hence the - sign}

and

f(1) =

lim f(x)
x -> 1+
{x approaches 1 from the right, hence the + sign}

Let's calculate these, shall we?

f(1) = 1^2 + b(1) + c|1 - 2| = 1 + b + c|-1| = 1 + b + c
We use the second line of the function because x is EQUAL to 1, and that function is used for x >= 1.

However,
lim f(x) = x + 1 {we take the FIRST function this time because}
x -> 1- {we are approaching from the LEFT}

lim f(x) = 1 + 1 = 2
x -> 1-

Therefore, we must equate:

1 + b + c = 2, or
b + c = 1

I'll leave it to you to finish the rest.

Answer 2

Interesting problem. Continuity simply means that the two functions have to be matched at x=1 and x=3.

From the first equation f(1)=2, f(3)=4

plug x=1 and x = 3 into the second equation and find values of b, c that match those f values.

x= 1: 1+b+c=2
x=3: 9+3b+c=4

b+c =1
3b+c= -5

b+c=1
2b=-6

b=-3, c=+4

Now check:

Puggy has the right idea: to not do all the work.

Answer 3

First the first line:
1 2 For the second line:
I x-2 I >=1 => 3 <= x & x<= 1 =>
x^2 +bx+c=3
x^2 +bx+c=1
then solve the equation and take the sharing limits.

Answer 4

First substitute x = 1. This gives 1 + b + c = 2
Next substitute x = 3. This gives 9 + 3b + c = 4
Solving these equations simultaneously gives b = -3 and c = 4