Answer to Question #264248 in Calculus for Artika

Question

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 .

Answer 1

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

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: $a=-\frac{61}{70},b=-\frac{41}{14}.$