Taylor Polynomials

It passes through $(a, f(a))$ with slope $f'(a)$.

• Correct: $y = f(a) + f'(a)(x - a)$:
• $y = f(a) + f'(x)(x - a)$:
• $y = f'(a)(x - a)$:
• $y = f(a)(x - a) + f'(a)$:

$f' = \cos x$, $f'' = -\sin x$, $f''' = -\cos x$.

• Correct: $-\cos x$:
• $\cos x$:
• $-\sin x$:
• $\sin x$:

$4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$.

• Correct: $24$:
• $16$:
• $10$:
• $4$:

The coefficient of $(x + 1)^2$ is $\frac{f''(-1)}{2!} = 3$, so $f''(-1) = 2 \cdot 3 = 6$.

• Answer: 6

The missing $(x - 3)$ term means $f'(3) = 0$, and $\frac{f''(3)}{2!} = -2$ gives $f''(3) = -4 < 0$. By the second derivative test, $f$ has a relative maximum at $x = 3$.

• $f$ has a relative minimum at $x = 3$: The coefficient of $(x - 3)^2$ is negative, so $f''(3) < 0$: concave down, a maximum.
• $f$ has a point of inflection at $x = 3$: $f''(3) = -4 \ne 0$, so the concavity is down at 3, not changing.
• $f$ is decreasing at $x = 3$: There is no $(x - 3)$ term, so $f'(3) = 0$.
• Correct: $f$ has a relative maximum at $x = 3$: Right: $f'(3) = 0$ and $f''(3) = -4 < 0$.

Every derivative of $e^x$ is $e^x$, which is 1 at $x = 0$, so $P_3(x)$$= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$, written with no "$+ \cdots$" since a polynomial has a last term.

• $1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$: A polynomial stops: the "$+ \cdots$" turns it into an endless sum.
• $x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}$: This drops $f(0) = 1$ and adds a fourth-degree term.
• Correct: $1 + x + \frac{x^2}{2} + \frac{x^3}{6}$: Right: four terms, through $x^3$, and it stops there.
• $1 + x + x^2 + x^3$: The coefficients need the factorials: $\frac{1}{2!}$ and $\frac{1}{3!}$.