Answer to Question #264248 in Calculus for Artika

Question #264248

Find the values of a and b that make f continuous everywhere.

f{x]= [(x^2+6x+5)/(x^2-3x-4)] , x < -1

(ax^2 - bx+ 3) , -1 less than equal to x < 3

2x - a + b , x Greater than equal to 3 .


1
Expert's answer
2021-11-11T19:00:41-0500

"\\frac{x^2+6x+5}{x^2-3x-4}=\\frac{(x+1)(x+5)}{(x+1)(x-4)}=\\frac{x+5}{x-4}"

If f(x) is continuous at x=-1, then "\\frac{-1+5}{-1-4}=a(-1)^2-b(-1)+3" or "-\\frac{4}{5}=a+b+3"

or "5a+5b=-19."

If f(x) is continuous at x=3, then "a(3^2 )-b(3)+3=2(3)-a+b" or "10a-4b=3."

So, to find the values of a and b that make f continuous everywhere, we have two equations:

"5a+5b=-19" and "10a-4b=3."

Solving these equation we have: "a=-\\frac{61}{70},b=-\\frac{41}{14}."


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