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?

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- Feb 19th 2010, 12:40 PMphalangedetermining 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.

so are the answers:

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

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

- Feb 19th 2010, 01:45 PMicemanfan
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.