Chapter 4.8, Problem 26E

To determine

To calculate: The function $f\left(x\right)$ from the equation $f^{\text{'}}\left(x\right)=2x-\frac{3}{x^4}$ with condition $f\left(1\right)=3,x>0$ .

Answer to Problem 26E

The function $f\left(x\right)=x^2+1x^{-3}+1$ from the equation $f^{\text{'}}\left(x\right)=2x-\frac{3}{x^4}$ .

Explanation of Solution

Given information:

The equation is given as:

  $f^{\text{'}}\left(x\right)=2x-\frac{3}{x^4}$ and condition $f\left(1\right)=3,x>0$ .

Formula used:

The function F is antiderivative of $f\left(x\right)$ on an interval $I$ if $F^{`}\left(x\right)=f\left(x\right)$ for all $x$ in $I$

The function $F$ is known as antiderivative of $f\left(x\right)$ on an interval $I$ the most general antiderivative of the function on interval $I$ is $F\left(x\right)+C$ , where $C$ is an arbitrary constant.

Calculation:

Consider the function,

  $f^{\text{'}}\left(x\right)=2x-\frac{3}{x^4}$

The antiderivative of $f\left(x\right)$ is,

  $F^{`}\left(x\right)=f\left(x\right)$

So, by reverse power rule Integrating both sides of equation,

  $\begin{array}{c}{\displaystyle \int F^{`}\left(x\right)}={\displaystyle \int f\left(x\right)}\\ F\left(x\right)={\displaystyle \int f\left(x\right)}\end{array}$

Therefore, integrating the function

  $\begin{array}{c}{\displaystyle \int f^{\text{'}}\left(x\right)}dx={\displaystyle \int 2x-\frac{3}{x^4}dx}\\ f\left(x\right)=2\frac{x^2}{2}-3\frac{x^{-4+1}}{-4+1}+C\\ f\left(x\right)=x^2+1x^{-3}+C\end{array}$

The value of $x$ is always $>0$ because if put $x=0$ then equation is undefined so value of $x>0$

Apply, the condition $f\left(1\right)=3,x>0$ on equation for obtain $C$ ,

  $\begin{array}{c}f\left(1\right)={\left(1\right)}^2+1{\left(1\right)}^{-3}+C\\ 3=1+1+C\\ 3-2=C\\ C=1\end{array}$

Put the value of $C$ in the equation,

  $f\left(x\right)=x^2+1x^{-3}+1$

Thus, the particular solution is $f\left(x\right)=x^2+1x^{-3}+1$ .