2.1 Use Taylor method to linearize f(x) = -3x^3 + 30x when x = -3.

2.1 Use Taylor method to linearize $f(x) = -3x^3 + 30x$ when $x = -3$. The Taylor linearization (first-order Taylor polynomial) of $f(x)$ around $x=a$ is given by $L(x) = f(a) + f'(a)(x-a)$. Here, $f(x) = -3x^3 + 30x$ and $a = -3$. Step 1: Calculate $f(a)$. $$f(-3) = -3(-3)^3 + 30(-3)$$ $$f(-3) = -3(-27) - 90$$ $$f(-3) = 81 - 90$$ $$f(-3) = -9$$ Step 2: Calculate the first derivative $f'(x)$. $$f'(x) = \frac{d}{dx}(-3x^3 + 30x)$$ $$f'(x) = -9x^2 + 30$$ Step 3: Calculate $f'(a)$. $$f'(-3) = -9(-3)^2 + 30$$ $$f'(-3) = -9(9) + 30$$ $$f'(-3) = -81 + 30$$ $$f'(-3) = -51$$ Step 4: Substitute $f(a)$ and $f'(a)$ into the linearization formula. $$L(x) = f(a) + f'(a)(x-a)$$ $$L(x) = -9 + (-51)(x - (-3))$$ $$L(x) = -9 - 51(x + 3)$$ $$L(x) = -9 - 51x - 153$$ $$L(x) = -51x - 162$$ The linearized function is $\boxed{-51x - 162}$. 2.2 If $R(x) = 22 + 25(x-2) + 11(x-2)^2 + 2(x-2)^3$, prove that $R(x)$ is the Taylor polynomial for $f(x) = 2x^3 - x^2 + 5x$, where $a = 2$. The Taylor polynomial of degree 3 for $f(x)$ centered at $a=2$ is given by: $$P_3(x) = f(2) + f'(2)(x-2) + \frac{f''(2)}{2!}(x-2)^2 + \frac{f'''(2)}{3!}(x-2)^3$$ We need to calculate the derivatives of $f(x)$ at $x=2$ and compare the coefficients. Given $f(x) = 2x^3 - x^2 + 5x$. Step 1: Calculate $f(2)$. $$f(2) = 2(2)^3 - (2)^2 + 5(2)$$ $$f(2) = 2(8) - 4 + 10$$ $$f(2) = 16 - 4 + 10$$ $$f(2) = 22$$ This matches the constant term in $R(x)$. Step 2: Calculate $f'(x)$ and $f'(2)$. $$f'(x) = 6x^2 - 2x + 5$$ $$f'(2) = 6(2)^2 - 2(2) + 5$$ $$f'(2) = 6(4) - 4 + 5$$ $$f'(2) = 24 - 4 + 5$$ $$f'(2) = 25$$ This matches the coefficient of $(x-2)$ in $R(x)$. Step 3: Calculate $f''(x)$ and $f''(2)$. $$f''(x) = 12x - 2$$ $$f''(2) = 12(2) - 2$$ $$f''(2) = 24 - 2$$ $$f''(2) = 22$$ The coefficient of $(x-2)^2$ in the Taylor polynomial is $\frac{f''(2)}{2!}$. $$\frac{f''(2)}{2!} = \frac{22}{2} = 11$$ This matches the coefficient of $(x-2)^2$ in $R(x)$. Step 4: Calculate $f'''(x)$ and $f'''(2)$. $$f'''(x) = 12$$ $$f'''(2) = 12$$ The coefficient of $(x-2)^3$ in the Taylor polynomial is $\frac{f'''(2)}{3!}$. $$\frac{f'''(2)}{3!} = \frac{12}{6} = 2$$ This matches the coefficient of $(x-2)^3$ in $R(x)$. Step 5: Form the Taylor polynomial $P_3(x)$ and compare it with $R(x)$. Substituting the calculated values: $$P_3(x) = 22 + 25(x-2) + 11(x-2)^2 + 2(x-2)^3$$ Since $P_3(x)$ is identical to $R(x)$, $R(x)$ is indeed the Taylor polynomial for $f(x)$ centered at $a=2$. 2.3 What makes you think that $R(x)$ is the Taylor polynomial of $f(x)$ in 2.2. $R(x)$ is the Taylor polynomial of $f(x)$ because its coefficients match the values of $f(x)$ and its successive derivatives evaluated at $x=2$, divided by the corresponding factorials. This directly follows the definition of a Taylor polynomial, where each coefficient $\frac{f^{(k)}(a)}{k!}$ was verified to be correct. Send me the next one 📸