Chapter G, Problem 46E

To determine

ToCalculate:The value of $f\text{'}\left(0\right)$

Answer to Problem 46E

The value of $f\text{'}\left(0\right)$ is $3$ .

Explanation of Solution

Given Information:

  $f$ is a quadratic function such that $f\left(0\right)=1$ and ${\displaystyle \int \frac{f\left(x\right)}{x^2{\left(x+1\right)}^3}}dx$

Calculation:

Here, $f\left(x\right)$ is quadratic function.

Assume that, $f\left(x\right)=a{x}^2+bx+c$

Since, $f\left(0\right)=1$

Substitute $x=0,$ in $f\left(x\right)=a{x}^2+bx+c$

  $\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\\ f\left(0\right)=a{\left(0\right)}^2+b\left(0\right)+c\\ 1=c\end{array}$

Substitute $c=1,$ in $f\left(x\right)=a{x}^2+bx+c$

Therefore, $f\left(x\right)=a{x}^2+bx+1$

Consider, ${\displaystyle \int \frac{f\left(x\right)}{x^2{\left(x+1\right)}^3}}dx$

  ${\displaystyle \int \frac{f\left(x\right)}{x^2{\left(x+1\right)}^3}}dx={\displaystyle \int \frac{a{x}^2+bx+1}{x^2{\left(x+1\right)}^3}}dx$

Degree of numerator $\left(a{x}^2+bx+1\right)$ is less than the degree of denominator $\left(x^2{\left(x+1\right)}^3\right)$

Factor $x$ is twice and $\left(x+1\right)$ is thrice, in denominator $\left(x^2{\left(x+1\right)}^3\right)$ .

Partial fraction decomposition of $\frac{a{x}^2+bx+1}{x^2{\left(x+1\right)}^3}$ as follows:

  $\begin{array}{l}\frac{a{x}^2+bx+1}{x^2{\left(x+1\right)}^3}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{\left(x+1\right)}+\frac{D}{{\left(x+1\right)}^2}+\frac{E}{{\left(x+1\right)}^3}\\ \frac{a{x}^2+bx+1}{x^2{\left(x+1\right)}^3}=\frac{Ax{\left(x+1\right)}^3+B{\left(x+1\right)}^3+C{x}^2{\left(x+1\right)}^2+D{x}^2\left(x+1\right)+E{x}^2}{x^2{\left(x+1\right)}^3}\\ a{x}^2+bx+1=Ax{\left(x+1\right)}^3+B{\left(x+1\right)}^3+C{x}^2{\left(x+1\right)}^2+D{x}^2\left(x+1\right)+E{x}^2\\ =Ax\left(x^3+3x^2+3x+1\right)+B\left(x^3+3x^2+3x+1\right)+C{x}^2\left(x^2+2x+1\right)+D{x}^2\left(x+1\right)+E{x}^2\\ =A{x}^4\left(A+C\right)+x^2\left(A+B+2C+D\right)+x^2\left(3A+3B+C+D+E\right)+x\left(A+3B\right)+B\end{array}$

Compare, $a{x}^2+bx+1$ with $x^4\left(A+C\right)+x^3\left(A+B+2C+D\right)+x^2\left(3A+3B+C+D+E\right)+x\left(A+3B\right)+B$

  $\begin{array}{l}A+C=0\\ 3A+B+2C+D=0\\ 3A+3B+C+D+E=a\\ a+3B=b\mathrm{....}(1)\\ B=1\mathrm{....}(2)\end{array}$

The objective is to evaluate $f\text{'}\left(0\right)$ .

Consider $f\left(x\right)=a{x}^2+bx+1$

Now, differentiate both sides of $f\left(x\right)=a{x}^2+bx+1$ , with respect to $x$ .

  $\begin{array}{l}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(a{x}^2+bx+1\right)\\ \frac{d}{dx}f\left(x\right)=2xa+b\end{array}$

Substitute, $x=0$ , in $f\text{'}\left(x\right)=2xa+b$

  $\begin{array}{l}f\text{'}\left(0\right)=2\left(0\right)a+b\\ f\text{'}\left(0\right)=b\mathrm{...}(3)\end{array}$

Consider, ${\displaystyle \int \frac{f\left(x\right)}{x^2{\left(x+1\right)}^3}dx}$

  $\begin{array}{l}{\displaystyle \int \frac{f\left(x\right)}{x^2{\left(x+1\right)}^3}dx}={\displaystyle \int \frac{a{x}^2+bx+1}{x^2{\left(x+1\right)}^3}}dx\\ ={\displaystyle \int \frac{A}{x}+}\frac{B}{x^2}+\frac{C}{\left(x+1\right)}+\frac{D}{{\left(x+1\right)}^2}+\frac{E}{{\left(x+1\right)}^3}dx\\ ={\displaystyle \int \frac{A}{x}dx+{\displaystyle \int \frac{B}{x^2}dx+{\displaystyle \int \frac{C}{\left(x+1\right)}dx+{\displaystyle \int \frac{D}{{\left(x+1\right)}^2}dx+{\displaystyle \int \frac{E}{{\left(x+1\right)}^3}dx}}}}}\end{array}$

Now recall that,

  ${\displaystyle \int \frac{1}{x+a}}dx=\log \left(x+a\right)$ and ${\displaystyle \int {\left(x-a\right)}^{n}dx=\frac{{\left(x+a\right)}^{n-1}}{n+1}}$

  $\begin{array}{l}{\displaystyle \int \frac{f\left(x\right)}{x^2{\left(x+1\right)}^3}dx}=A{\displaystyle \int \frac{1}{x}}dx+B{\displaystyle \int x^{2}dx+C{\displaystyle \int \frac{1}{\left(x+1\right)}}}dx+D{\left(x+1\right)}^{-2}dx+E{\displaystyle \int {\left(x+1\right)}^{-3}dx}\\ =A\log \left(x\right)+B\left(\frac{x^{-1}}{-1}\right)+C\log \left(x+1\right)+D\left(\frac{{\left(x+1\right)}^{-3}}{-1}\right)+E\left(\frac{{\left(x+1\right)}^{-2}}{-2}\right)\\ =A\log \left(x\right)-\frac{B}{x}+C\log \left(x+1\right)-\frac{D}{\left(x+1\right)}-\frac{E}{2{\left(x+1\right)}^2}\end{array}$

It is given that ${\displaystyle \int \frac{f\left(x\right)}{x^2{\left(x+1\right)}^3},dx}$ is rational function.

But $\log \left(x\right)$ is not rational function.

Hence, $A\log \left(x\right)-\frac{B}{x}+C\log \left(x+1\right)-\frac{D}{\left(x+1\right)}-\frac{E}{2{\left(x+1\right)}^2}$

Constants $A$ and $C$ must be zero to make the function

  $A\log \left(x\right)-\frac{B}{x}+C\log \left(x+1\right)-\frac{D}{\left(x+1\right)}-\frac{E}{2{\left(x+1\right)}^2}$ , a rational function.

It implies that $A=0$ … $\left(4\right)$

Substitute, $A=0$ from equation $\left(4\right)$ and $B=1$ from equation $\left(2\right)$ , in equation $\left(1\right)$ .

  $\begin{array}{l}A+3B=b\\ 0+3\left(1\right)=b\\ b=3\end{array}$

Substitute, $b=3$ in equation $\left(3\right)$

  $\begin{array}{l}f\text{'}\left(0\right)=b\\ f\text{'}\left(0\right)=3\end{array}$