Taylor Polynomial Approximation - AP Calculus BC

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$\begin{align}&$\text{For the function values: }$\&f(-2)=2;f'(-2)=10;f''(-2)=-3;f'''(-2)=-3\&$\text{Approximate }$f(-1)$\text{ using a Taylor expansion.}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(-2)=2;f'(-2)=10;f''(-2)=-3;f'''(-2)=-3\&$\text{ at }$x=-1\&$\text{This expansion becomes:}$\&f(-1)\approx $$\frac{2}{0!}$(-1-(-2))^{0}$$+\frac{10}{1!}$(-1-(-2))^{1}$$+\frac{-3}{2!}$(-1-(-2))^{2}$$+\frac{-3}{3!}$(-1-(-2))^{3}$\&f(-1)\approx10\end{align}$

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Question

$\begin{align}&$\text{For the function values: }$\&f(-12)=6;f'(-12)=19;f''(-12)=18;f'''(-12)=-2\&$\text{Approximate }$f(-10)$\text{ using a Taylor expansion.}$\end{align}$

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$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(-12)=6;f'(-12)=19;f''(-12)=18;f'''(-12)=-2\&$\text{ at }$x=-10\&$\text{This expansion becomes:}$\&f(-10)\approx $$\frac{6}{0!}$(-10-(-12))^{0}$$+\frac{19}{1!}$(-10-(-12))^{1}$$+\frac{18}{2!}$(-10-(-12))^{2}$$+\frac{-2}{3!}$(-10-(-12))^{3}$\&f(-10)\approx$\frac{232}{3}$\end{align}$

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Question

$\begin{align}&$\text{For the function values: }$\&f(9)=10;f'(9)=-16;f''(9)=7;f'''(9)=-2;f''''(9)=-12\&$\text{Approximate }$f(6)$\text{ using a Taylor expansion.}$\end{align}$

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$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(9)=10;f'(9)=-16;f''(9)=7;f'''(9)=-2;f''''(9)=-12\&$\text{ at }$x=6\&$\text{This expansion becomes:}$\&f(6)\approx $$\frac{10}{0!}$(6-9)^{0}$$+\frac{-16}{1!}$(6-9)^{1}$$+\frac{7}{2!}$(6-9)^{2}$$+\frac{-2}{3!}$(6-9)^{3}$$+\frac{-12}{4!}$(6-9)^{4}$\&f(6)\approx58\end{align}$

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Question

$\begin{align}&$\text{Approximate }$f(7) \&$\text{Where }$f(3)=7;f'(3)=16;f''(3)=1\&$\text{With a Taylor series expansion.}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(3)=7;f'(3)=16;f''(3)=1\&$\text{ at }$x=7\&$\text{This expansion becomes:}$\&f(7)\approx $$\frac{7}{0!}$(7-3)^{0}$$+\frac{16}{1!}$(7-3)^{1}$$+\frac{1}{2!}$(7-3)^{2}$\&f(7)\approx79\end{align}$

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Question

$\begin{align}&$\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.}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\end{align}$

$\begin{align}\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(-2)=7;f'(-2)=-19;f''(-2)=13;f'''(-2)=10;f''''(-2)=-16\&$\text{ at }$x=-7\&$\text{This expansion becomes:}$\&f(-7)\approx $$\frac{7}{0!}$(-7-(-2))^{0}$$+\frac{-19}{1!}$(-7-(-2))^{1}$$+\frac{13}{2!}$(-7-(-2))^{2}$$+\frac{10}{3!}$(-7-(-2))^{3}$$+\frac{-16}{4!}$(-7-(-2))^{4}$\&f(-7)\approx-$\frac{721}{2}$\end{align}$

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Question

$\begin{align}&$\text{Approximate }$f(-4) \&$\text{Where }$f(-3)=-3;f'(-3)=-13;f''(-3)=-10;f'''(-3)=-20\&$\text{With a Taylor series expansion.}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(-3)=-3;f'(-3)=-13;f''(-3)=-10;f'''(-3)=-20\&$\text{ at }$x=-4\&$\text{This expansion becomes:}$\&f(-4)\approx $$\frac{-3}{0!}$(-4-(-3))^{0}$$+\frac{-13}{1!}$(-4-(-3))^{1}$$+\frac{-10}{2!}$(-4-(-3))^{2}$$+\frac{-20}{3!}$(-4-(-3))^{3}$\&f(-4)\approx$\frac{25}{3}$\end{align}$

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Question

$\begin{align}&$\text{Approximate }$f(3) \&$\text{Where }$f(8)=17;f'(8)=12;f''(8)=3\&$\text{By using a Taylor expansion.}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(8)=17;f'(8)=12;f''(8)=3\&$\text{ at }$x=3\&$\text{This expansion becomes:}$\&f(3)\approx $$\frac{17}{0!}$(3-8)^{0}$$+\frac{12}{1!}$(3-8)^{1}$$+\frac{3}{2!}$(3-8)^{2}$\&f(3)\approx-$\frac{11}{2}$\end{align}$

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Question

$\begin{align}&$\text{Approximate }$f(-5) \&$\text{Where }$f(-2)=-18;f'(-2)=11;f''(-2)=7\&$\text{With a Taylor series expansion.}$\end{align}$

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$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(-2)=-18;f'(-2)=11;f''(-2)=7\&$\text{ at }$x=-5\&$\text{This expansion becomes:}$\&f(-5)\approx $$\frac{-18}{0!}$(-5-(-2))^{0}$$+\frac{11}{1!}$(-5-(-2))^{1}$$+\frac{7}{2!}$(-5-(-2))^{2}$\&f(-5)\approx-$\frac{39}{2}$\end{align}$

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Question

$\begin{align}&$\text{Approximate }$f(3) \&$\text{Where }$f(2)=13;f'(2)=-7;f''(2)=13;f'''(2)=12\&$\text{With a Taylor series expansion.}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(2)=13;f'(2)=-7;f''(2)=13;f'''(2)=12\&$\text{ at }$x=3\&$\text{This expansion becomes:}$\&f(3)\approx $$\frac{13}{0!}$(3-2)^{0}$$+\frac{-7}{1!}$(3-2)^{1}$$+\frac{13}{2!}$(3-2)^{2}$$+\frac{12}{3!}$(3-2)^{3}$\&f(3)\approx$\frac{29}{2}$\end{align}$

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Question

$\begin{align}&$\text{Approximate }$f(6) \&$\text{Where }$f(11)=-18;f'(11)=15;f''(11)=5;f'''(11)=-6\&$\text{With a Taylor series expansion.}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(11)=-18;f'(11)=15;f''(11)=5;f'''(11)=-6\&$\text{ at }$x=6\&$\text{This expansion becomes:}$\&f(6)\approx $$\frac{-18}{0!}$(6-11)^{0}$$+\frac{15}{1!}$(6-11)^{1}$$+\frac{5}{2!}$(6-11)^{2}$$+\frac{-6}{3!}$(6-11)^{3}$\&f(6)\approx$\frac{189}{2}$\end{align}$

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Question

$\begin{align}&$\text{For the function values: }$\&f(1)=3;f'(1)=-13;f''(1)=19;f'''(1)=-10;f''''(1)=17\&$\text{Approximate }$f(-3)$\text{ using a Taylor expansion.}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(1)=3;f'(1)=-13;f''(1)=19;f'''(1)=-10;f''''(1)=17\&$\text{ at }$x=-3\&$\text{This expansion becomes:}$\&f(-3)\approx $$\frac{3}{0!}$(-3-1)^{0}$$+\frac{-13}{1!}$(-3-1)^{1}$$+\frac{19}{2!}$(-3-1)^{2}$$+\frac{-10}{3!}$(-3-1)^{3}$$+\frac{17}{4!}$(-3-1)^{4}$\&f(-3)\approx495\end{align}$

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Question

$\begin{align}&$\text{Approximate }$f(-5) \&$\text{Where }$f(-8)=-11;f'(-8)=12;f''(-8)=20;f'''(-8)=-19\&$\text{By using a Taylor expansion.}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function values }$f(-8)=-11;f'(-8)=12;f''(-8)=20;f'''(-8)=-19\&$\text{ at }$x=-5\&$\text{This expansion becomes:}$\&f(-5)\approx $$\frac{-11}{0!}$(-5-(-8))^{0}$$+\frac{12}{1!}$(-5-(-8))^{1}$$+\frac{20}{2!}$(-5-(-8))^{2}$$+\frac{-19}{3!}$(-5-(-8))^{3}$\&f(-5)\approx$\frac{59}{2}$\end{align}$

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Question

$\begin{align}&$\text{Approximate }$f(6) \&$\text{Where }$f(x)=-sin(4x)sin(5x)\&$\text{With a Taylor series centered at }$2\cdot \pi$\text{, using }$3$\text{ terms}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function }$-sin(4x)sin(5x)$\text{ at }$x=6\&$\text{This expansion becomes:}$\&f(x)\approx (-sin(4a)sin(5a))+(- 4cos(4a)sin(5a) - $5cos(5a)sin(4a))(x-a)^{1}$+\frac{41sin(4a)sin(5a) - $40cos(4a)cos(5a)}{2}$(x-a)^{2}$\&x=6\&a=2\cdot \pi\&f(6)\approx-1.60\end{align}$

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Question

$\begin{align}&$\text{Approximate }$f(1) \&$\text{Where }$f(x)=- sin(6x) - sin(7x)\&$\text{With a Taylor series centered at }$$\frac{(15\cdot \pi )}{4}$$\text{, using }$3$\text{ terms}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function }$- sin(6x) - sin(7x)$\text{ at }$x=1\&$\text{This expansion becomes:}$\&f(x)\approx (- sin(6a) - sin(7a))+(- 6cos(6a) - $7cos(7a))(x-a)^{1}$+\frac{36sin(6a) + $49sin(7a)}{2}$(x-a)^{2}$\&x=1\&a=\frac{(15\cdot \pi )}{4}$\&f(1)\approx4157.36\end{align}$

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Question

$\begin{align}&$\text{For the function: }$f(x)=sin(9x)sin(x)\&$\text{Approximate }$f(4)\&$\text{Using a Taylor expansion centered at }$$\frac{(15\cdot \pi )}{4}$$\text{ to }$2$\text{ terms}$\end{align}$

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$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function }$sin(9x)sin(x)$\text{ at }$x=4\&$\text{This expansion becomes:}$\&f(x)\approx (sin(9a)sin(a))+(9cos(9a)sin(a) + $sin(9a)cos(a))(x-a)^{1}$\&x=4\&a=\frac{(15\cdot \pi )}{4}$\&f(4)\approx39.40\end{align}$

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Question

$\begin{align}&$\text{Approximate }$f(10) \&$\text{Where }$f(x)=\frac{cos(5x)}{sin(5x)}$\&$\text{By using a Taylor expansion to }$2$\text{ terms, centered at }$$\frac{\pi }{2}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function }$$\frac{cos(5x)}{sin(5x)}$$\text{ at }$x=10\&$\text{This expansion becomes:}$\&f(x)\approx ($\frac{cos(5a)}{sin(5a)}$)+(-$\frac{ $(5cos(5a)^{2}$$$)}{sin(5a)^{2}$}- $5)(x-a)^{1}$\&x=10\&a=\frac{\pi }{2}$\&f(10)\approx-42.15\end{align}$

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Question

$\begin{align}&$\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}$\end{align}$

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Answer

$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\end{align}$

$\begin{align}\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function }$cos(x) +\frac{ 1}{cos(8x)}$$\text{ at }$x=5\&$\text{This expansion becomes:}$\&f(x)\approx (cos(a) +\frac{ $1}{cos(8a)}$)+($\frac{(8sin(8a))}{cos(8a)^{2}$$}- $sin(a))(x-a)^{1}$+\frac{\frac{64}{cos(8a)}$- cos(a) +\frac{ $(128sin(8a)^{2}$$$)}{cos(8a)^{3}$$}}{2}(x-a)^{2}$\&x=5\&a=-4\cdot \pi\&f(5)\approx9722.19\end{align}$

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Question

$\begin{align}&$\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}$\end{align}$

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$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function }$-cos(6x)sin(2x)$\text{ at }$x=9\&$\text{This expansion becomes:}$\&f(x)\approx (-cos(6a)sin(2a))+(6sin(2a)sin(6a) - $2cos(2a)cos(6a))(x-a)^{1}$\&x=9\&a=-$\frac{(3\cdot \pi )}{2}$\&f(9)\approx-27.42\end{align}$

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Question

$\begin{align}&$\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}$\end{align}$

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$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function }$-$\frac{cos(8x)}{cos(4x)}$$\text{ at }$x=9\&$\text{This expansion becomes:}$\&f(x)\approx (-$\frac{cos(8a)}{cos(4a)}$)+($\frac{(8sin(8a))}{cos(4a)}$-$\frac{ $(4cos(8a)sin(4a))}{cos(4a)^{2}$$$})(x-a)^{1}$\&x=9\&a=\frac{(5\cdot \pi )}{4}$\&f(9)\approx1.00\end{align}$

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Question

$\begin{align}&$\text{For the function: }$f(x)=sin(2x) - cos(7x)\&$\text{Approximate }$f(7)\&$\text{Using a Taylor expansion centered at }$-2\cdot \pi$\text{ to }$3$\text{ terms}$\end{align}$

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$\begin{align}&$\text{The Taylor series is a means of approximating a function value}$\&$\text{at given point, if values for the function and its derivatives}$\&$\text{are known at another point. It takes advantage of rates of change}$\&$\text{(i.e. function derivative values), and the distance between}$\&$\text{points on the x-axis to approximate changes along the y-axis.}$\&$\text{It is similar to using the slope of a line to find changes in}$\&$\text{y with regards to changes in x. The Taylor series follows the}$\&$\text{form:}$\&f(x)\approx\sum_${i=0}^{n}$$$\frac{f^{(i)}$$$(a)}{i!}(x-a)^i$\approx $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}$(x-a)^2$$+\frac{f^{(3)}$$$(a)}{3!}(x-a)^3$$+...+\frac{f^{(n)}$$$(a)}{n!}(x-a)^n$\&$\text{For the function }$sin(2x) - cos(7x)$\text{ at }$x=7\&$\text{This expansion becomes:}$\&f(x)\approx (sin(2a) - cos(7a))+(2cos(2a) + $7sin(7a))(x-a)^{1}$+\frac{49cos(7a) - $4sin(2a)}{2}$(x-a)^{2}$\&x=7\&a=-2\cdot \pi\&f(7)\approx4348.42\end{align}$

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