Answer to Question #264268 in Algebra for ella

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<=x<3),(2x-a+b,x>=3):}

Answer 1

f (x)={(( x²+6x+5)/(x²-3x-4),x <-1), ax²-bx+3,-1 <x <3),(2x-a+b,x>=3)}

x²+ 6x+ 5= x²+ x+ 5x+ 5

=x (x+1) + 5 (x+1)

= (x+5) (x+1)

x²- 3x- 4= x²- 4x+ x- 4

x (x-4)+1 (x-4)= (x+1) (x-4)

Therefore, (x+5) (x+1)/ (x+1) (x-4)

=(x+5)/ (x-4)

x cannot be equal to 4 since the denominator will equate to zero.

Taking x= -1; f (x)=(-1+ 5) / (-1- 4)

=-4/ 5

lim_x~-1 ax²- bx+ 3= -4/ 5

a+ b=-19/5, let this be equation 1.

lim_x~3 ax²-bx+3=-4/5

9a-3b=-19/5, let this be equation 2.

lim_x~3 2x- a+ b= -4/5

-a+ b= -34/5, let this be equation 3.

Taking equation 1 and 2;

9a-3b= -19/5

a+b= -19/5

9a-3b= -19/5

9a+27b= -171/5; substracting the 2 equations;

-30b= 152/5

b= 76/75

a+b=-19/5

a+76/75=-19/5

a+76/75=-19/5

a=-361/75

a=-361/75

b=76/75