Chapter 4.8, Problem 36E

To determine

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

Answer to Problem 36E

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

Explanation of Solution

Given information:

The function is given as:

  $f^{\text{'}\text{'}\text{'}}\left(x\right)=\cos x$ andconditions $f\left(0\right)=1,f^{\text{'}}\left(0\right)=2,f^{\text{'}\text{'}\text{'}}\left(0\right)=3$ .

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,

  $f^{\text{'}\text{'}\text{'}}\left(x\right)=\cos x$

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

  $\begin{array}{c}{f}^{\text{'}\text{'}\text{'}}\left(x\right)=\cos x\\ {\displaystyle \int f^{\text{'}\text{'}\text{'}}\left(x\right)}dx={\displaystyle \int \cos x}dx\\ f^{\text{'}\text{'}}\left(x\right)=\sin x+C\end{array}$

Again, integrate the function,

  $\begin{array}{l}{\displaystyle \int f^{\text{'}\text{'}}\left(x\right)dx}={\displaystyle \int \sin x+C}dx\\ f^{\text{'}}\left(x\right)=-\cos x+Cx+D\end{array}$

Again, integrate the function,

  $\begin{array}{l}{\displaystyle \int f^{\text{'}}\left(x\right)}dx={\displaystyle \int -\cos x+Cx+D}dx\\ f\left(x\right)=-\sin x+C\frac{x^2}{2}+Dx+E\end{array}$

Put $f\left(0\right)=1$ in the equation $f\left(x\right)=-\sin x+C\frac{x^2}{2}+Dx+E$ ,

  $\begin{array}{c}f\left(0\right)=-\sin 0+C\frac{0^2}{2}+D0+E\\ E=1\end{array}$

Put $f^{\text{'}}\left(0\right)=2$ in the equation $f^{\text{'}}\left(x\right)=-\cos x+Cx+D$ ,

  $\begin{array}{c}{f}^{\text{'}}\left(0\right)=-\cos 0+C0+D\\ D=3\end{array}$

Put $f^{\text{'}\text{'}}\left(0\right)=3$ in the equation $f^{\text{'}\text{'}}\left(x\right)=\sin x+C$

  $\begin{array}{c}{f}^{\text{'}\text{'}}\left(0\right)=\sin 0+C\\ C=3\end{array}$

Put the values of $C$ , $E$ and $D$ in equation $f\left(x\right)=-\sin x+C\frac{x^2}{2}+Dx+E$

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

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