Find a and b analytically, so that f(x) is continous everywhere.?

Question

f(x) =

{x²-1, x<-1
{ax+b, -1 {rad(2x+5), x>2

Answer 1

So find a and b such that there is no discontinuity at -1 or 2
i.e. Lim x->h- f(x) = Lim x->h+ f(x)

Two points: in your definition, f(x) is undefined at -1 and 2. There is a "<=" missing from the domain of either one of the piecewise functions at each point. So strictly it only has a limiting definition at these points.
Also, not sure what function rad() means, did you mean 'radical' as in square-root? (a cube-root is also a radical, and so on).

Anyway:
Lim x-> -1- [f(x)] = Lim x->-1- [x²-1] = 0
Lim x-> -1+ [f(x)] = Lim x->-1+ ax+b = (b-a)
Therefore continuity at x->-1 requires b-a = 0, i.e. a=b

Lim x-> 2- [f(x)] = Lim x->2- ax+b = 2a+b
Lim x-> 2+ [f(x)] = Lim x->2+ rad(2x+5) = rad(9)? , whatever that means
Therefore continuity at x->2 requires 2a+b = 3a = rad(9) =3
a = rad(9)/3 =1

Solution: a=1, b=1

Answer 2

f(-1) = 0 = b - a
f(2) = 3 = 2a + b
b = a = 1

1(-1) + 1 = 0
1(2) + 1 = 3

Answer 3

to be continuous, at the end of the intervals, the two functions need to intersect..

so set up some simultaneous equations and solve..

Also, if they need to be differential...then the derivatives also need to be equal at the ends of the intervals.