lnx,x>=1

          lnx,x>=1

           /

f(x)={

           \

              e^(x-1)+βx-β-1,x<1

f has a derivative at x=1

find b

If there is a derivative at x=1 then the curve must be continuous and smooth at x=1

\(f(1) = ln(1)=0\\ f(x) = lnx \qquad for \;\;x\ge1\\ f ' (x) = \frac{1}{x} \qquad for \;\;x\ge1\\ f ' (1) = \frac{1}{1}=1 \)

Now I am going to ignore the fact that the following function is not defined for x=1

but it needs to have the same f(1) and f ' (1) as    f(x)=ln x does

\(f(x)=e^{(x-1)}+\beta x-\beta-1\\ f(x)=\frac{e^x}{e}+\beta x-\beta-1\\ f'(x)=e^x+\beta\\ NOW\\ f(1)=0\\ f(1)=e^{(1-1)}+\beta* 1-\beta-1\\ f(1)=1+\beta-\beta-1\\ f(1)=0 \qquad good \\also\\ f'(1)=1\\ f'(1)=e^1+\beta\\ 1=e+\beta\\ \beta=1-e \)

I do not think I have displayed this very well but  beta = 1-e

Thanks!!!