e^(x^2-x+a) if x < 1
as x->1-, f(x) -> e^a
(4^x-x^2)) if 1<x<2
as x->1+, f(x) -> 4^1-1 = 3
as x->2-, f(x) -> 4^2-2^2 = 12
6In(x-b) if x>2
as x->2-, f(x) -> 6ln(2-b)
so, we need
e^a = 3
a = ln3
6ln(2-b) = 12
ln(2-b)=2
2-b = e^2
b = 2-e^2
But using those values will not make f(x) continuous at every x, since f(x) is not defined at x=1 or x=2. To fix that, we need to add
f(1) = 3
f(2) = 12
Or make some of the intervals closed on one end or or maybe both ends.
Determine the values of a and b to make the following function continuous at every value of x.?
f(x)= { e^(x^2-x+a) if x < 1 , (4^x-x^2)) if 1<x<2 , 6In(x-b) if x>2
1 answer