\begin{align*} f(x) & = \ln (2x^2 + 3) \\ \\ f'(x) & = {4x \over 2x^2 + 3} \phantom{000000} \left[ {d \over dx} [ \ln f(x) ] = {f'(x) \over f(x)} \right] \\ \\ f'(x) & > 0 \phantom{0000000000000} [\text{Increasing function}] \\ {4x \over 2x^2 + 3} & > 0 \\ \\ \text{For all } & \text{real values of } x , \\ x^2 & \ge 0 \\ 2x^2 & \ge 0 \\ 2x^2 + 3 & \ge 3 \\ \\ \therefore 4x & > 0 \\ x & > 0 \end{align*}

\begin{align*} y & = \ln (e^{2x} + 1) \\ \\ {dy \over dx} & = { 2e^{2x} \over e^{2x} + 1} \phantom{000000} \left[ {d \over dx} [ \ln f(x) ] = {f'(x) \over f(x)} \text{ and } {d \over dx} [ e^{f(x)} ] = f'(x) e^{f(x)} \right] \\ \\ \text{For all} & \text{ real values of } x, e^{2x} > 0 \\ \implies & {2e^{2x} \over e^{2x} + 1} >0 \\ \implies & \phantom{00} {dy \over dx} > 0 \\ \\ \therefore y = \ln (e^{2x} + 1) & \text{ is an increasing function for all real values of } x \end{align*}

\begin{align*} y & = -x^3 + 3x^2 - 5x + 1 \\ \\ {dy \over dx} & = - 3x^2 + 3(2)(x) - 5 \\ & = -3x^2 + 6x - 5 \\ & = - 3 (x^2 - 2x) - 5 \\ & = - 3 \left[ x^2 - 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 \right] - 5 \phantom{000000} [\text{Complete the square}] \\ & = -3 [ (x - 1)^2 - 1 ] - 5 \\ & = - 3(x - 1)^2 + 3 - 5 \\ & = - 3(x - 1)^2 - 2 \end{align*} \begin{align*} \text{For all real } & \text{values of } x , \\ (x - 1)^2 & \ge 0 \\ -3(x - 1)^2 & \le 0 \\ -3(x - 1)^2 - 2 & \le -2 \\ \\ \implies {dy \over dx} < 0 & \text{ for all real values of } x \\ \\ \therefore y \text{ is a decreasing } & \text{function for all real values of } x \end{align*}