$ f(x)= \begin{cases} e^{a+bx}&\text{ }\, 0\leq x\ < \frac{1}{2}\\ \mu &\text{ }\, \frac{1}{2} \leq x\leq 1\\ \end{cases} $
ASSUME $\mu$ is a constant and a and b are the two parameters
I was wondering how I would go about making this piecewise function continuous.
Ideas: I believe that I have to try and find the values of the two parameters
1) I would let $x=\frac{1}{2}$ then $$ e^{a+\frac{1}{2}b} = \mu $$ $$a+\frac{1}{2}b = log(\mu) $$ $$a= log(\mu) -\frac{1}{2}b $$ But I'm not sure how to continue from here.
2) Begin by taking the limits of either side $\frac{1}{2}$
$$\lim_{x \to \frac{1}{2}+} f(x) = \lim_{x \to \frac{1}{2}+}e^{a+bt}= e^{a+\frac{1}{2}b}$$ $$\lim_{x \to \frac{1}{2}-} f(x) = \mu $$
This implies that: $$ e^{a+\frac{1}{2}b} = \mu $$ This will eventually bring me back to the dilemma in idea 1.
Is either my 1st or 2nd method correct in dealing with the problem, if so could somebody kindly show me how to follow through to solve it and if not could somebody show me the correct path to take, so that I can tackle the problem ?
Let's start from the very basic. Say you have a piecewise defined function and in it's definition at see point say c the function is not continous. The reasons for this might be 1) lhl≠rhl≠f(c) 2) either limit is not defined or is oscillatory. 3) lhl=rhl≠f(c) Case 3 can be tackled easily by setting up lhl = rhl= f(c) by redefining the function at c . But in your problem you have to deal with two variables and have only one equation and you could just paremetrize your ordered pairs of (a,b) into one variable and nothing more.