91Ó°ÊÓ

We have the given function $f\left(x\right)=\sin \left(3x\right)$with a derivative of order $4$ at$x=\frac{\mathrm{Ï€}}{6}$ .

The fourth taylor polynomial for $x=\frac{\mathrm{Ï€}}{6}$is given by,

$\begin{array}{rl}{P}_4\left(x\right)=& f\left(\frac{\mathrm{Ï€}}{6}\right)+f^{\mathrm{′}}\left(\frac{\mathrm{Ï€}}{6}\right)(x−\frac{\mathrm{Ï€}}{6})+\frac{f^{′â\P IJ}\left(\frac{\mathrm{Ï€}}{6}\right)}{2!}{(x−\frac{\mathrm{Ï€}}{6})}^2+\frac{f^{′â\P IJ}\left(\frac{\mathrm{Ï€}}{6}\right)}{3!}{(x−\frac{\mathrm{Ï€}}{6})}^3+\frac{f^{′â\P IJ′â\P IJ}\left(\frac{\mathrm{Ï€}}{6}\right)}{4!}{(x−\frac{\mathrm{Ï€}}{6})}^4\end{array}$

Therefore, we have to find the value of the function along with $f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right),f^{\text{'}\text{'}\text{'}}\left(x\right)$and $f\text{'}\text{'}\text{'}\text{'}\left(x\right)$at $x=\frac{\mathrm{Ï€}}{6}$.

The value of the function at $x=\frac{\mathrm{Ï€}}{6}$is,

$\begin{array}{rl}f\left(\frac{\mathrm{Ï€}}{6}\right)& =\sin (3â‹\dots \frac{\mathrm{Ï€}}{6})\\ & =\sin \left(\frac{\mathrm{Ï€}}{2}\right)\\ & =1\end{array}$

The derivatives of the function, $f\left(x\right)=\sin \left(3x\right)$

$\begin{array}{r}{f}^{\mathrm{′}}\left(x\right)=\frac{d}{dx}\left[\sin 3x\right]\\ =3\cos 3x\end{array}$

So, at $x=\frac{\mathrm{Ï€}}{6}$

$\begin{array}{rl}{f}^{\mathrm{′}}\left(\frac{\mathrm{Ï€}}{6}\right)& =3\cos (3â‹\dots \frac{\mathrm{Ï€}}{6})\\ & =3\cos \left(\frac{\mathrm{Ï€}}{2}\right)\\ & =3â‹\dots 0\\ & =0\end{array}$

$\begin{array}{rl}{f}^{′â\P IJ}\left(x\right)& =\frac{d}{dx}\left[3\cos 3x\right]\\ & =3\frac{d}{dx}\left[\cos 3x\right]\\ & =3(−3\sin 3x)\\ & =−9\sin 3x\end{array}$

So, at $x=\frac{\mathrm{Ï€}}{6}$

$\begin{array}{rl}{f}^{′â\P IJ}\left(\frac{\mathrm{Ï€}}{6}\right)& =−9\sin (3â‹\dots \frac{\mathrm{Ï€}}{6})\\ & =−9\sin \left(\frac{\mathrm{Ï€}}{2}\right)\\ & =−9â‹\dots 1\\ & =−9\end{array}$

$\begin{array}{rl}{f}^{′â\P IJ\mathrm{′}}\left(x\right)& =−9\frac{d}{dx}\left[\sin 3x\right]\\ & =−9\left(3\cos 3x\right)\\ & =−27\cos 3x\end{array}$

So, at $x=\frac{\mathrm{Ï€}}{6}$

$\begin{array}{rl}{f}^{′â\P IJ\mathrm{′}}\left(\frac{\mathrm{Ï€}}{6}\right)& =−27\cos (3â‹\dots \frac{\mathrm{Ï€}}{6})\\ & =−27\cos \left(\frac{\mathrm{Ï€}}{2}\right)\\ & =−27â‹\dots 0\\ & =0\end{array}$

$\begin{array}{rl}{f}^{′â\P IJ′â\P IJ}\left(x\right)& =\frac{d}{dx}[−27\cos 3x]\\ & =−27\frac{d}{dx}\left[\cos 3x\right]\\ & =−27(−3\sin 3x)\\ & =81\sin 3x\end{array}$

So, at $x=\frac{\mathrm{Ï€}}{6}$

$\begin{array}{rl}{f}^{′â\P IJ′â\P IJ}\left(\frac{\mathrm{Ï€}}{6}\right)& =81\sin (3â‹\dots \frac{\mathrm{Ï€}}{6})\\ & =81\sin \left(\frac{\mathrm{Ï€}}{2}\right)\\ & =81â‹\dots 1\\ & =81\end{array}$

Hence the fourth Taylor polynomial of the function $f\left(x\right)=\sin 3x$at$\frac{\mathrm{Ï€}}{6}$

$\begin{array}{rl}{P}_4\left(x\right)& =1+0â‹\dots (x−\frac{\mathrm{Ï€}}{6})+\frac{(−9)}{2!}{(x−\frac{\mathrm{Ï€}}{6})}^2+\frac{0}{3!}{(x−\frac{\mathrm{Ï€}}{6})}^3+\frac{81}{4!}{(x−\frac{\mathrm{Ï€}}{6})}^4\\ & =1−\frac{9}{2!}{(x−\frac{\mathrm{Ï€}}{6})}^2+\frac{81}{4!}{(x−\frac{\mathrm{Ï€}}{6})}^4\\ & =1−\frac{9}{2}{(x−\frac{\mathrm{Ï€}}{6})}^2+\frac{27}{8}{(x−\frac{\mathrm{Ï€}}{6})}^4\end{array}$