# determining a and b

• Feb 19th 2010, 12:40 PM
phalange
determining a and b
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.

a=-2 and b=2?
• Feb 19th 2010, 12:57 PM
icemanfan
Is the question:

What are the values of a and b that make f(x) a continuous function?
• Feb 19th 2010, 01:37 PM
phalange
yes. sorry i forgot to add that
• Feb 19th 2010, 01:45 PM
icemanfan
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.