2.1 Use Taylor method to linearize f(x) = -3x³ + 30x when x = -3.

$$f(-3) = -3(-3)^3 + 30(-3)$$ $$f(-3) = -3(-27) - 90$$ $$f(-3) = 81 - 90$$ $$f(-3) = -9$$

$$f'(x) = \frac{d}{dx}(-3x^3 + 30x)$$ $$f'(x) = -9x^2 + 30$$

$$f'(-3) = -9(-3)^2 + 30$$ $$f'(-3) = -9(9) + 30$$ $$f'(-3) = -81 + 30$$ $$f'(-3) = -51$$

$$L(x) = f(a) + f'(a)(x-a)$$ $$L(x) = -9 + (-51)(x - (-3))$$ $$L(x) = -9 - 51(x + 3)$$ $$L(x) = -9 - 51x - 153$$ $$L(x) = -51x - 162$$

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 $f(x) = 2x^3 - x^2 + 5x$.

$$f(2) = 2(2)^3 - (2)^2 + 5(2)$$ $$f(2) = 2(8) - 4 + 10$$ $$f(2) = 16 - 4 + 10$$ $$f(2) = 22$$

This matches the constant term in $R(x)$.

$$f'(x) = 6x^2 - 2x + 5$$ $$f'(2) = 6(2)^2 - 2(2) + 5$$ $$f'(2) = 6(4) - 4 + 5$$ $$f'(2) = 24 - 4 + 5$$ $$f'(2) = 25$$

This matches the coefficient of $(x-2)$ in $R(x)$.

$$f''(x) = 12x - 2$$ $$f''(2) = 12(2) - 2$$ $$f''(2) = 24 - 2$$ $$f''(2) = 22$$

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 $R(x)$.

$$f'''(x) = 12$$ $$f'''(2) = 12$$

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 $R(x)$.

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.