2.1 Use Taylor method to linearize $f(x) = -3x^3 + 30x$ when $x = -3$. Step 1: Find $f(a)$ and $f'(x)$. The linearization of $f(x)$ at $x=a$ is given by $L(x) = f(a) + f'(a)(x-a)$. Given $f(x) = -3x^3 + 30x$ and $a = -3$. First, calculate $f(a)$: $$ f(-3) = -3(-3)^3 + 30(-3) = -3(-27) - 90 = 81 - 90 = -9 $$ Next, find the first derivative $f'(x)$: $$ f'(x) = \frac{d}{dx}(-3x^3 + 30x) = -9x^2 + 30 $$ Step 2: Calculate $f'(a)$. Substitute $a=-3$ into $f'(x)$: $$ f'(-3) = -9(-3)^2 + 30 = -9(9) + 30 = -81 + 30 = -51 $$ Step 3: Form the linearization $L(x)$. Substitute $f(-3)$ and $f'(-3)$ into the linearization formula: $$ L(x) = f(-3) + f'(-3)(x - (-3)) $$ $$ L(x) = -9 + (-51)(x + 3) $$ $$ L(x) = -9 - 51x - 153 $$ $$ L(x) = -51x - 162 $$ The linearization of $f(x)$ at $x=-3$ is: $$ \boxed{L(x) = -51x - 162} $$ 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 $a = 2$. To prove $R(x)$ is the Taylor polynomial for $f(x)$ at $a=2$, we need to show that the coefficients of $R(x)$ match the Taylor series coefficients $\frac{f^{(k)}(a)}{k!}$. The Taylor polynomial of degree 3 centered at $a$ is: $$ P_3(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 $$ Given $f(x) = 2x^3 - x^2 + 5x$ and $a=2$. Step 1: Calculate $f(a)$. $$ f(2) = 2(2)^3 - (2)^2 + 5(2) = 2(8) - 4 + 10 = 16 - 4 + 10 = 22 $$ This matches the constant term in $R(x)$. Step 2: Calculate $f'(a)$. First, find the first derivative $f'(x)$: $$ f'(x) = \frac{d}{dx}(2x^3 - x^2 + 5x) = 6x^2 - 2x + 5 $$ Now, evaluate $f'(2)$: $$ f'(2) = 6(2)^2 - 2(2) + 5 = 6(4) - 4 + 5 = 24 - 4 + 5 = 25 $$ This matches the coefficient of $(x-2)$ in $R(x)$. Step 3: Calculate $f''(a)$. First, find the second derivative $f''(x)$: $$ f''(x) = \frac{d}{dx}(6x^2 - 2x + 5) = 12x - 2 $$ Now, evaluate $f''(2)$: $$ f''(2) = 12(2) - 2 = 24 - 2 = 22 $$ The Taylor coefficient for $(x-a)^2$ is $\frac{f''(a)}{2!}$: $$ \frac{f''(2)}{2!} = \frac{22}{2} = 11 $$ This matches the coefficient of $(x-2)^2$ in $R(x)$. Step 4: Calculate $f'''(a)$. First, find the third derivative $f'''(x)$: $$ f'''(x) = \frac{d}{dx}(12x - 2) = 12 $$ Now, evaluate $f'''(2)$: $$ f'''(2) = 12 $$ The Taylor coefficient for $(x-a)^3$ is $\frac{f'''(a)}{3!}$: $$ \frac{f'''(2)}{3!} = \frac{12}{6} = 2 $$ This matches the coefficient of $(x-2)^3$ in $R(x)$. Since all calculated Taylor coefficients $f(2)$, $f'(2)$, $\frac{f''(2)}{2!}$, and $\frac{f'''(2)}{3!}$ match the coefficients in $R(x)$, $R(x)$ 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. The reason $R(x)$ is considered the Taylor polynomial of $f(x)$ in 2.2 is that its coefficients are precisely the Taylor coefficients of $f(x)$ evaluated at $a=2$. Specifically, for each term $(x-2)^k$ in $R(x)$, its coefficient is equal to $\frac{f^{(k)}(2)}{k!}$, where $f^{(k)}(2)$ is the $k$-th derivative of $f(x)$ evaluated at $x=2$. That's 2 down. 3 left today — send the next one.