Answer
$f(x)=-\ln(x)+\ln(2)x-\ln(2)$
Work Step by Step
$$f''(x)=x^{-2}$$
$$\int f''(x)dx=\int x^{-2} dx$$
$$f'(x)=\frac{x^{-1}}{-1}+c$$
$$f'(x)=-x^{-1}+c$$
$$\int f'(x)dx=\int (-x^{-1}+c)dx$$
$$f(x)=\int (-x^{-1}+c)dx$$
$$f(x)=\int (-\frac{1}{x}+c)dx$$
$$f(x)=-\ln(x)+cx+c_{1}$$
Find the constant using the given conditions.
$$\begin{cases}0=-\ln(1)+c\cdot 1+c_{1} \\ 0=-\ln(2)+c\cdot 2+c_{1} \end{cases}$$
$$\begin{cases}0=0+c+c_{1} \\ 0=-\ln(2)+2c+c_{1} \end{cases}$$
$$\begin{cases}-c=c_{1} \\ 0=-\ln(2)+2c+c_{1} \end{cases}$$
$$\begin{cases}-c=c_{1} \\ 0=-\ln(2)-2c_{1}+c_{1} \end{cases}$$
$$\begin{cases}-c=c_{1} \\ 0=-\ln(2)-c_{1} \end{cases}$$
$$\begin{cases}-c=c_{1} \\ -\ln(2)=c_{1} \end{cases}$$
$$\begin{cases}-c=-\ln(2) \\ -\ln(2)=c_{1} \end{cases}$$
$$\begin{cases}c=\ln(2) \\ -\ln(2)=c_{1} \end{cases}$$
$$f(x)=-\ln(x)+\ln(2)x-\ln(2)$$
