Question

Find the values of a and b that make f continuous everywhere.
f(x) =
(x^2-9)/(x-3) if x<3
ax^2-bx+3 if 3<=x<4
2x-a+b if x>=4

Answer 1

the two answers i got were, b=-1, a=0. what i did was set:
(x^2-9)/(x-3)=ax^2-bx+3 and from this i got b=3a-1, and then i set ax^2-bx+3=2x-a+b.

Answer 2

find limit
left hand limit of x->3
lim x->3 (x+3(x-3)/(x-3)=6
if it has to be continuous then left and right limits must be equal .
right hand limits of x->3

a(3)^2-3b+3=6
9a-3b=3

now left and right limits of x->4

left hand limit
16a-4b+3
right hand limit
8-a+b
so they should be equal
16a-4b+3=8-a+b
17a-5b=5

solve the following two equations
9a-3b=3
17a-5b=5
which results the same
a=0
b=-1
you got it correct !

Answer 3

okay tyvm