For the piecewise function f(x) = a - 10x for x < sin(e); f(x) = bx^2 + 2 for x greater or equal to sin(e) how do you need to choose a(b) such that f(x) is continuos at x = sin(e)

Question

In order to solve this I made both parts of the piecewise function equal, substituted x for sin(e) and solved for a. I got the answer a(b) = b*sin^2(e) + 10sin(e) + 2. However, I am unsure if my method was correct and would like clarification.

Answer 1

Hi Karim,

Yes you are correct... the reason is, as stated, there are really two functions one for x<sin(e) and another for x≥sin(e). In order to make the two equal at x= sin(e) take the limit as x→sin(e) of both functions and then set them equal solve for a as a function of b i.e. a(b) that's what you did. So once e and b are specified a can be calculated so that f(x) is continuous at x=sin(e).

Hope this helps

Jim