f(x)= 2x-1. 1<x<3
f(x)= x^2+ax+b, lx-2l greater than or equal to 1
so i know that the absolute value means x is greater than or equal to 3 and x is also less than or equal to 1.
so are the answers:
a=-2 and b=2?
Ok, then you need the limit of the function to be the same when x approaches 1 from the left and from the right, and also the same when x approaches 3 from the left and from the right.
This means
$\displaystyle 2x - 1 = x^2 + ax + b$ when x = 1 and when x = 3.
Substituting those values of x into the equation, we get these two equations:
$\displaystyle 2(1) - 1 = 1^2 + a(1) + b$
$\displaystyle 2(3) - 1 = 3^2 + a(3) + b$
Simplifying, we get
$\displaystyle a + b = 0$
$\displaystyle 3a + b = -4$
$\displaystyle a = -2, b = 2$
You are correct.