Question

Please show me how to do this.

Answer 1

Answer:

$f(x) = \dfrac{1}{2}(x - \ln (x - 1) - 2)$

Step-by-step explanation:

$f'(x) = \dfrac{1}{2} - \dfrac{1}{2x - 2}$

$f(x) = \int (\dfrac{1}{2} - \dfrac{1}{2x - 2}) dx$

$f(x) = \int [\dfrac{1}{2} - \dfrac{1}{2(x - 2)}] dx$

$f(x) = \int \dfrac{1}{2}dx - \dfrac{1}{2} \int \dfrac{1}{x - 1} dx$

$f(x) = \dfrac{1}{2}x - \dfrac{1}{2} \ln (x - 1) + C$

$f(x) = \dfrac{x - \ln (x - 1)}{2} + C$

$f(2) = \dfrac{2 - \ln (2 - 1)}{2} + C = 0$

$f(2) = \dfrac{2 - \ln (1)}{2} + C = 0$

$f(2) = \dfrac{2}{2} + C = 0$

$f(2) = 1 + C = 0$

$C = -1$

$f(x) = \dfrac{x - \ln (x - 1)}{2} - 1$

$f(x) = \dfrac{1}{2}[x - \ln (x - 1) - 2]$