In Exercises 41–48 find the fourth Taylor polynomial $P_4\left(x\right)$for the specified function and the given value of $x_0$.

$\text{48.}\sin 3x,\frac{\pi }{6}$

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

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

$\begin{array}{rl}{P}_4\left(x\right)=& f\left(\frac{\pi }{6}\right)+f^{\mathrm{\prime }}\left(\frac{\pi }{6}\right)(x-\frac{\pi }{6})+\frac{f^{\prime \prime }\left(\frac{\pi }{6}\right)}{2!}{(x-\frac{\pi }{6})}^2+\frac{f^{\prime \prime }\left(\frac{\pi }{6}\right)}{3!}{(x-\frac{\pi }{6})}^3+\frac{f^{\prime \prime \prime \prime }\left(\frac{\pi }{6}\right)}{4!}{(x-\frac{\pi }{6})}^4\end{array}$

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 $x=\frac{\mathrm{\pi }}{6}$.

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

$\begin{array}{rl}f\left(\frac{\pi }{6}\right)& =\sin (3\cdot \frac{\pi }{6})\\ & =\sin \left(\frac{\pi }{2}\right)\\ & =1\end{array}$

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

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

So, at $x=\frac{\pi }{6}$

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(\frac{\pi }{6}\right)& =3\cos (3\cdot \frac{\pi }{6})\\ & =3\cos \left(\frac{\pi }{2}\right)\\ & =3\cdot 0\\ & =0\end{array}$

$\begin{array}{rl}{f}^{\prime \prime }\left(x\right)& =\frac{d}{dx}\left[3\cos 3x\right]\\ & =3\frac{d}{dx}\left[\cos 3x\right]\\ & =3(-3\sin 3x)\\ & =-9\sin 3x\end{array}$

So, at $x=\frac{\pi }{6}$

$\begin{array}{rl}{f}^{\prime \prime }\left(\frac{\pi }{6}\right)& =-9\sin (3\cdot \frac{\pi }{6})\\ & =-9\sin \left(\frac{\pi }{2}\right)\\ & =-9\cdot 1\\ & =-9\end{array}$

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

So, at $x=\frac{\pi }{6}$

$\begin{array}{rl}{f}^{\prime \prime \mathrm{\prime }}\left(\frac{\pi }{6}\right)& =-27\cos (3\cdot \frac{\pi }{6})\\ & =-27\cos \left(\frac{\pi }{2}\right)\\ & =-27\cdot 0\\ & =0\end{array}$

$\begin{array}{rl}{f}^{\prime \prime \prime \prime }\left(x\right)& =\frac{d}{dx}[-27\cos 3x]\\ & =-27\frac{d}{dx}\left[\cos 3x\right]\\ & =-27(-3\sin 3x)\\ & =81\sin 3x\end{array}$

So, at $x=\frac{\pi }{6}$

$\begin{array}{rl}{f}^{\prime \prime \prime \prime }\left(\frac{\pi }{6}\right)& =81\sin (3\cdot \frac{\pi }{6})\\ & =81\sin \left(\frac{\pi }{2}\right)\\ & =81\cdot 1\\ & =81\end{array}$

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

$\begin{array}{rl}{P}_4\left(x\right)& =1+0\cdot (x-\frac{\pi }{6})+\frac{(-9)}{2!}{(x-\frac{\pi }{6})}^2+\frac{0}{3!}{(x-\frac{\pi }{6})}^3+\frac{81}{4!}{(x-\frac{\pi }{6})}^4\\ & =1-\frac{9}{2!}{(x-\frac{\pi }{6})}^2+\frac{81}{4!}{(x-\frac{\pi }{6})}^4\\ & =1-\frac{9}{2}{(x-\frac{\pi }{6})}^2+\frac{27}{8}{(x-\frac{\pi }{6})}^4\end{array}$