Chapter 7, Problem 50RE

To determine

To find: The function $f$ that has a continuous derivative on $\left(0,\infty \right)$ .

Answer to Problem 50RE

The function $f$ is $f\left(x\right)=\frac{1}{4}{x}^2-\frac{1}{2}\ln \left|x\right|+\frac{3}{4}$ .

Explanation of Solution

Given information:

The graph of $f$ goes through the point $\left(1,1\right)$ .

The given formula is $L=\ln x+f\left(x\right)-1$ .

Calculation:

If $f$ is a continuous function on $\left[a,b\right]$ ,

  $F\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)}$

The derivative of $F\left(x\right)$ is,

  $\begin{array}{c}{F}^{\text{'}}\left(x\right)=\frac{d}{dx}\left[{\displaystyle {\int }_a^{x}f\left(t\right)}dt\right]\\ =f\left(x\right)\end{array}$

The length of the curve is from $t=1\text{ and }t=x$ .

  $L\left(x\right)={\displaystyle {\int }_1^x\sqrt{1+{\left[f\text{'}\left(t\right)\right]}^{2}dt}}$

Differentiating the above equation to get,

  $L\text{'}\left(x\right)=\sqrt{1+{\left[f\text{'}\left(t\right)\right]}^2}$

It is given that,

  $L=\ln x+f\left(x\right)-1$

Differentiating to get,

  $L\text{'}\left(x\right)=\frac{1}{x}+f\text{'}\left(x\right)-0$

Now relating the above equations,

  $\sqrt{1+{\left[f\text{'}\left(t\right)\right]}^2}=\frac{1}{x}+f\text{'}\left(x\right)-0$

Squaring on both sides to get,

  $\begin{array}{c}1+{\left[f\text{'}\left(t\right)\right]}^2=\frac{1}{x^2}+{\left[f\text{'}\left(x\right)\right]}^2+\frac{2f\text{'}\left(x\right)}{x}\\ 1-\frac{1}{x^2}=\frac{2f\text{'}\left(x\right)}{x}\end{array}$

Multiplying on both sides by $\frac{x}{2}$ to get,

  $\frac{x}{2}-\frac{1}{2x}=f\text{'}\left(x\right)$

Integrating on both sides,

  $\begin{array}{c}{\displaystyle \int \frac{x}{2}-\frac{1}{2x}dx}=f\left(x\right)\\ f\left(x\right)=\frac{x^2}{4}-\frac{1}{2}\ln \left|x\right|+C\end{array}$

To find $C$ replacing $x$ and $f\left(x\right)$ by $1$ ,

  $1=\frac{1}{4}-\frac{1}{2}\ln 1+C$

Subtract on both sides by $\frac{1}{4}$ to get,

  $C=\frac{3}{4}$

Substituting the value of $C$ ,

  $f\left(x\right)=\frac{1}{4}{x}^2-\frac{1}{2}\ln \left|x\right|+\frac{3}{4}$

Therefore, the function $f$ that has a continuous derivative on $\left(0,\infty \right)$ is $f\left(x\right)=\frac{1}{4}{x}^2-\frac{1}{2}\ln \left|x\right|+\frac{3}{4}$ .