Taylor Polynomial Approximation - AP Calculus BC

Card 0 of 240

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Tap the card to reveal the answer