Given continuous function f(x)=1, x<=0, f(x) =2^xa, 0<x<=1, f(x) = (x^2-2x+b)/(-x+2), 1<x<=2, what is b^(a+2)?

domain of f (-infinite,2)

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Given `f(x)={[1,x<=0],[2^x a,0<x<=1],[(x^2-2x+b)/(-x+2),1<x<2]}`

where f(x) is continuous, find the values of a and b.

** There is a typo in the problem. f(x) cannot be continuous at x=2 for any values of a or b since there is a vertical asymptote at x=2 for all values of a or b. The domain should be `(-oo,2)` so the inequality should be strict for x<2. **

For a function to be continuous it must be continuous at every point on its domain. Since each part of the piecewise function is everywhere continuous, we need only look at x=0,x=1 since these are the endpoints of the intervals.

Now a function is continuous at a point if the function is defined at the point, the limit exists at the point and the value of the function agrees with the limit. f(x) is defined at x=0 (f(0)=1) and defined at x=1 ` `(f(1)=2a ) so we need to check the limits.

For a limit of a function to exist at a point the left and right limits must agree.

`lim_(x->0^-)=1` since the function is constant on `(-oo,0)`

`lim_(x->0^+)=2^0 a=a` (The exponential function is continuous everywhere so we can use substitution to find the limit.)

**So if f(x) is continuous, the limits must agree and a=1.**

`lim_(x->1^-)=2^1 a=2a=2` substituting the known value of a.

`lim_(x->1^+)=((1)^2-2(1)+b)/(-(1)+2)=-1+b` (Again the rational function is continuous everywhere except x=2, so we can use substitution at x=1 to find the limit.)

**So for f(x) to be continuous the limits must agree and -1+b=2 ==> b=3.**

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**Since f(x) is continuous a=1,b=3 and `b^(a+2)=3^3=27` **

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The graph with a=1,b=3:

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