Answer:
[tex] f(x) = \dfrac{1}{2}(x - \ln (x - 1) - 2) [/tex]
Step-by-step explanation:
[tex] f'(x) = \dfrac{1}{2} - \dfrac{1}{2x - 2} [/tex]
[tex] f(x) = \int (\dfrac{1}{2} - \dfrac{1}{2x - 2}) dx [/tex]
[tex] f(x) = \int [\dfrac{1}{2} - \dfrac{1}{2(x - 2)}] dx [/tex]
[tex] f(x) = \int \dfrac{1}{2}dx - \dfrac{1}{2} \int \dfrac{1}{x - 1} dx [/tex]
[tex] f(x) = \dfrac{1}{2}x - \dfrac{1}{2} \ln (x - 1) + C [/tex]
[tex] f(x) = \dfrac{x - \ln (x - 1)}{2} + C [/tex]
[tex] f(2) = \dfrac{2 - \ln (2 - 1)}{2} + C = 0 [/tex]
[tex] f(2) = \dfrac{2 - \ln (1)}{2} + C = 0 [/tex]
[tex] f(2) = \dfrac{2}{2} + C = 0 [/tex]
[tex] f(2) = 1 + C = 0 [/tex]
[tex] C = -1 [/tex]
[tex] f(x) = \dfrac{x - \ln (x - 1)}{2} - 1 [/tex]
[tex] f(x) = \dfrac{1}{2}[x - \ln (x - 1) - 2] [/tex]
