Taylor Polynomial Approximation

Practice Questions

AP Calculus BC › Taylor Polynomial Approximation

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$\begin{align*}&\text{Approximate }f(6) \&\text{Where }f(x)=- cos(9x) - cos(x)\&\text{With a Taylor series centered at }-\frac{(5\cdot \pi )}{2}\text{, using }3\text{ terms}\end{align*}$

$\begin{align*}&\text{For the function: }f(x)=cos(x) +\frac{ 1}{cos(8x)}\&\text{Approximate }f(5)\&\text{Using a Taylor expansion centered at }-4\cdot \pi\text{ to }3\text{ terms}\end{align*}$

$\begin{align*}&\text{Approximate }f(9) \&\text{Where }f(x)=-\frac{cos(8x)}{cos(4x)}\&\text{By using a Taylor expansion to }2\text{ terms, centered at }\frac{(5\cdot \pi )}{4}\end{align*}$

$\begin{align*}&\text{Approximate }f(5) \&\text{Where }f(x)=sin(6x) - cos(2x)\&\text{By using a Taylor expansion to }3\text{ terms, centered at }\pi\end{align*}$

$\begin{align*}&\text{For the function: }f(x)=-cos(4x)^{2}\&\text{Approximate }f(3)\&\text{Using a Taylor expansion centered at }6\cdot \pi\text{ to }2\text{ terms}\end{align*}$

$\begin{align*}&\text{Approximate }f(1) \&\text{Where }f(x)=- cos(5x) - sin(2x)\&\text{With a Taylor series centered at }\frac{(5\cdot \pi )}{2}\text{, using }3\text{ terms}\end{align*}$

$\begin{align*}&\text{For the function: }f(x)=-cos(6x)sin(2x)\&\text{Approximate }f(9)\&\text{Using a Taylor expansion centered at }-\frac{(3\cdot \pi )}{2}\text{ to }2\text{ terms}\end{align*}$

$\begin{align*}&\text{Approximate }f(-7) \&\text{Where }f(-2)=7;f'(-2)=-19;f''(-2)=13;f'''(-2)=10;f''''(-2)=-16\&\text{By using a Taylor expansion.}\end{align*}$