Section+3.8

These are the notes from class on Tuesday, November 1st.

Today, we expanded on the Taylor Expansion formula and found new ways for finding the tangent line at the intersection of a function. We started off with a problem that Ms. Tyson made up. Let math $ f(x)=5x^{4}-20x^{3}+3x^{2}-1 $ math

Using the Taylor Expansion formula, find the equation of the tangent line at x = -2. The form would be math $ f(x)=A(x+2)^{4}+B(x+2)^{3}+C(x+2)^{2}+D(x+2)+E $ math where math $ y=D(x+2)+E $ math would be the equation of the tangent line.

Use repeated synthetic division.

Then, you would get math $ f(x)=5(x+2)^{4}-60(x+2)^{3}+243(x+2)^{2}-412(x+2)+251 $ math Therefore, the equation of the tangent line would be math $ y=-412(x+2)+251 $ math math $ y=-412x-824+251 $ math math $ y=-412x-573 $ math

After we solved the same problem using the method of undetermined coefficients, according to the textbook. math $ f(x)=5x^{4}-20x^{3}+3x^{2}-1=A+B(x+2)+C(x)(x+2)^{2} $ math where the equation of the tangent line would be math $ y=A+B(x+2) $ math

Plug in x=-2. math $ f(-2)=5(-2)^{4}-20(-2)^{3}+3(-2)^{2}-1=A+B(0)+C(-2)(0)^{2}=A $ math math $ A=251 $ math

Then, math $ f(x)=251+B(x+2)+C(x)(x+2)^{2} $ math Move A to the left side. math $ f(x)=5x^{4}-20x^{3}+3x^{2}-252=B(x+2)+C(x)(x+2)^{2} $ math

Use repeated synthetic division.

You get math $ (x+2)(5x^{3}-30x^{2}-126)=B(x+2)+C(x)(x+2)^{2} $ math math $ 5x^{3}-30x^{2}-126=B(x+2)+C(x)(x+2) $ math

Plug in x=-2. math $ 5(-2)^{3}-30(-2)^{2}-126=B(0)+C(-2)(0) $ math math $ B=-412 $ math

Therefore, the equation of the tangent line is math $ y=251-412(x+2) $ math Notice how we get the same answer that we got when we used the regular method.

The undetermined coefficients method is more complicated than the Taylor Expansion formula method. Therefore, it should be used only when dealing with rational functions, as the Taylor Expansion works only for polynomial functions.

Then, we practiced more with using the undetermined coefficients method. Let math $ f(x)=\frac{x^{3}-3x^{2}}{x-4} $ math Find the equation of the tangent line at x=1. math $ f(x)=A+B(x-1)+C(x)(x-1)^{2} $ math The equation of the tangent line is math $ y=A+B(x-1) $ math

Plug in x=1. math $ f(1)=\frac{1^{3}-3(1)^{2}}{1-4}=A+B(0)+C(1)(0)^{2} $ math math $ A=\frac{1-3}{1-4}=\frac{-2}{-3}=\frac{2}{3} $ math

Then, move A to the other side. math $ f(x)=\frac{x^{3}-3x^{2}}{x-4}-\frac{2}{3}=B(x-1)+C(x)(x-1)^{2} $ math math $ \frac{3x^{3}-9x^{2}}{3(x-4)}-\frac{2(x-4)}{3(x-4)}=(x-1)(B+C(x)(x-1) $ math math $ \frac{3x^{3}-9x^{2}-2x+8}{3(x-4)}=(x-1)(B+C(x)(x-1) $ math

If (x-1) is the factor of the left hand side, then (x-1) is the factor of its numerator. Use repeated synthetic division.

Therefore, you get math $ f(x)=\frac{(x-1)(3x^{2}-6x-8)}{3(x-4)}=(x-1)(B+C(x)(x-1)) $ math math $ f(x)=\frac{3x^{2}-6x-8}{3(x-4)}=B+C(x)(x-1) $ math

Plug in x=1. math $ f(x)=\frac{3(1)-6(1)-8}{3(-3)}=B+C(1)(0) $ math math B=\frac{-11}{-9}=\frac{11}{9} $ math

Then, plugging in the values for A and B, math $ f(x)=\frac{2}{3}+\frac{11}{9}(x-1)+C(x)(x-1)^{2} $ math

The equation of the tangent would therefore be math $ y=\frac{2}{3}+\frac{11}{9}(x-1) $ math

You can also check this on Wolfram Alpha, by typing taylor expansion of (x^3-3x^2)(x-4) at x=1