Exercise 6

Throughout our work in Section, we have focused on approximating the function $f(x) = e^x$. In this exercise, we change the function of interest to $f(x) = \frac{1}{3}x^3 + \frac{1}{4}x^2 - 2x - 1$, and consider the linear and quadratic approximations to $f$ near $a = 0$.

1. Determine $f'(x)$ and $f''(x)$ and enter their formulas below. $$\begin{aligned} f(x) &= \frac{1}{3}x^3 + \frac{1}{4}x^2 - 2x - 1 \\ f'(x) &= \underline{\hspace{2em}} \\ f''(x) &= \underline{\hspace{2em}} \end{aligned}$$

2. Next, compute $f(0)$, $f'(0)$, and $f''(0)$ and enter those values below. $$\begin{aligned} f(0) &= \underline{\hspace{2em}} \\ f'(0) &= \underline{\hspace{2em}} \\ f''(0) &= \underline{\hspace{2em}} \end{aligned}$$

3. Use your work so far to determine the formula for $T_1(x)$, the tangent line approximation to $f(x)$ at $a =0$ (which satisfies $T_1(0) = f(0)$ and $T_1'(0) = f'(0)$).

4. Let $T_2(x)$ be the quadratic approximation to $f(x)$ near $a = 0$ that satisfies $T_2(0) = f(0)$, $T_2'(0) = f'(0)$, and $T_2''(0) = f''(0)$. You might start by letting $T_2(x) = c_0 + c_1 x + c_2 x^2$, and creating an updated table like the one shown below.

   $$\begin{aligned} f(x) &= x^3/3 + x^2/4 - 2x - 1 \\ f'(x) &= \underline{\hspace{2em}} \\ f''(x) &= \underline{\hspace{2em}} \\ & \\ f(0) &= \underline{\hspace{2em}} \\ f'(0) &= \underline{\hspace{2em}} \\ f''(0) &= \underline{\hspace{2em}} \end{aligned}$$

   $$\begin{aligned} T_2(x) &= c_0 + c_1 x + c_2 x^2 \\ T_2'(x) &= c_1 + 2c_2 x \\ T_2''(x) &= \underline{\hspace{2em}} \\ & \\ T_2(0) &= \underline{\hspace{2em}} \\ T_2'(0) &= \underline{\hspace{2em}} \\ T_2''(0) &= \underline{\hspace{2em}} \end{aligned}$$

   Use your work in in the table above to find the formula for $T_2(x)$ that satisfies $T_2(0) = f(0)$, $T_2'(0) = f'(0)$, and $T_2''(0) = f''(0)$.

5. Plot $f(x)$, $T_1(x)$, and $T_2(x)$ on the same axes, centered at $a = 0$. What do you notice?