\(f(x)=\ln x\)

\(f'(x)=\frac{1}{x}\)

\(f''(x)=-\frac{1}{x^2}\)

Plug into the curvature formula

\(k(x)=\frac{|f''(x)|}{[1+(f'(x))^2]^\frac{3}{2}}=\frac{|-\frac{1}{x^2}|}{[1+(\frac{1}{x})^2]^\frac{3}{2}}=\frac{1}{x^2[1+\frac{1}{x^2}]^\frac{3}{2}}\)

Find k'

\(k(x)=\frac{1}{x^2[1+\frac{1}{x^2}]^{3/2}}=x^{-2}[1+\frac{1}{x^2}]^{-3/2}\)

\(k'(x)=-2x^{-3}[1+\frac{1}{x^2}]^{3/2}+x^{-2}\cdot-\frac{3}{2}[1+\frac{1}{x^2}]^{-5/2}(-2x^{-3})\)

\(=\frac{-2}{x^3[1+\frac{1}{x^2}]^{3/2}}+\frac{3}{x^5[1+\frac{1}{x^2}]^{5/2}}\)

\(=\frac{-2x^2(1+\frac{1}{x^2})+3}{x^5[1+\frac{1}{x^2}]^{5/2}}\)

\(=\frac{-2x^2-2+3}{x^5[1+\frac{1}{x^2}]^{5/2}}\)

\(=\frac{-2-2x^2}{x^5[1+\frac{1}{x^2}]^{5/2}}\)

Find where k'=0

\(0=1-2x^2\)

\(2x^2=1\)

\(x=\frac{1}{\sqrt{2}}\approx0.7071\)

The negative root is not in the domain of ln x. We can check numbers in the surrounding intervals to make sure it is the max:

\((0,0,7):\quad k'(0,1)\approx0.96\)

\((0,7,\infty):\quad k'(1)\approx-0.18\)

k is increasing then decreasing, so it is the max.

As \(x\to\infty\):

\(\lim_{x\to\infty}\frac{1}{x^2[1+\frac{1}{x^2}]^{3/2}}\to\frac{1}{\infty[1+0]^{3/2}}\to0\)

the curvature approaches 0.

\(f'(x)=\frac{1}{x}\)

\(f''(x)=-\frac{1}{x^2}\)

Plug into the curvature formula

\(k(x)=\frac{|f''(x)|}{[1+(f'(x))^2]^\frac{3}{2}}=\frac{|-\frac{1}{x^2}|}{[1+(\frac{1}{x})^2]^\frac{3}{2}}=\frac{1}{x^2[1+\frac{1}{x^2}]^\frac{3}{2}}\)

Find k'

\(k(x)=\frac{1}{x^2[1+\frac{1}{x^2}]^{3/2}}=x^{-2}[1+\frac{1}{x^2}]^{-3/2}\)

\(k'(x)=-2x^{-3}[1+\frac{1}{x^2}]^{3/2}+x^{-2}\cdot-\frac{3}{2}[1+\frac{1}{x^2}]^{-5/2}(-2x^{-3})\)

\(=\frac{-2}{x^3[1+\frac{1}{x^2}]^{3/2}}+\frac{3}{x^5[1+\frac{1}{x^2}]^{5/2}}\)

\(=\frac{-2x^2(1+\frac{1}{x^2})+3}{x^5[1+\frac{1}{x^2}]^{5/2}}\)

\(=\frac{-2x^2-2+3}{x^5[1+\frac{1}{x^2}]^{5/2}}\)

\(=\frac{-2-2x^2}{x^5[1+\frac{1}{x^2}]^{5/2}}\)

Find where k'=0

\(0=1-2x^2\)

\(2x^2=1\)

\(x=\frac{1}{\sqrt{2}}\approx0.7071\)

The negative root is not in the domain of ln x. We can check numbers in the surrounding intervals to make sure it is the max:

\((0,0,7):\quad k'(0,1)\approx0.96\)

\((0,7,\infty):\quad k'(1)\approx-0.18\)

k is increasing then decreasing, so it is the max.

As \(x\to\infty\):

\(\lim_{x\to\infty}\frac{1}{x^2[1+\frac{1}{x^2}]^{3/2}}\to\frac{1}{\infty[1+0]^{3/2}}\to0\)

the curvature approaches 0.