Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For f(x) = √(1+x) centered at x=0, the constant term is f(0) = 1, the linear coefficient is f'(0)/1! = 1/2, the quadratic coefficient is f''(0)/2! = -1/4 / 2 = -1/8, and the cubic coefficient is f'''(0)/3! = 3/8 / 6 = 1/16. These derivatives are f'(x) = $(1/2)(1+x)^{-1/2}$, f''(x) = $(1/2)(-1/2)(1+x)^{-3/2}$ = -1/4 $(1+x)^{-3/2}$, and f'''(x) = -1/4 $(-3/2)(1+x)^{-5/2}$ = 3/8 $(1+x)^{-5/2}$, evaluated at x=0. This results in 1 + (1/2)x - $(1/8)x^2$ + $(1/16)x^3$. A tempting distractor like choice D fails by using incorrect coefficients, such as -1/4 for quadratic instead of -1/8 after dividing by 2!. Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.

Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For f(x) = 1/(1-x) centered at x=0, the constant term is f(0) = 1, the linear coefficient is f'(0)/1! = 1, the quadratic coefficient is f''(0)/2! = 2/2 = 1, the cubic coefficient is f'''(0)/3! = 6/6 = 1, and the quartic coefficient is f''''(0)/4! = 24/24 = 1. These derivatives increase factorially as f'(x) = $1/(1-x)^2$, f''(x) = $2/(1-x)^3$, f'''(x) = $6/(1-x)^4$, and f''''(x) = $24/(1-x)^5$, evaluated at x=0. This yields 1 + x + $x^2$ + $x^3$ + $x^4$. A tempting distractor like choice B fails by alternating signs, which applies to 1/(1+x) instead of 1/(1-x). Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.

Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For f(x) = cos x centered at x=0, the constant term is f(0) = 1, the linear coefficient is f'(0)/1! = 0, the quadratic coefficient is f''(0)/2! = -1/2, the cubic coefficient is f'''(0)/3! = 0, and the quartic coefficient is f''''(0)/4! = 1/24. These derivatives are f'(x) = -sin x, f''(x) = -cos x, f'''(x) = sin x, and f''''(x) = cos x, evaluated at x=0. This results in 1 - $(x^2$)/2 + $(x^4$)/24. A tempting distractor like choice D fails by incorrectly making the quartic term negative, ignoring the positive fourth derivative. Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.

Constructing Taylor polynomials involves approximating a function using its derivatives at a center point. For f(x) = ln x centered at x=1, the constant term is f(1) = 0, the linear coefficient is f'(1)/1! = 1, the quadratic coefficient is f''(1)/2! = -1/2, and the cubic coefficient is f'''(1)/3! = 2/6 = 1/3. These derivatives are f'(x) = 1/x, f''(x) = $-1/x^2$, and f'''(x) = $2/x^3$, evaluated at x=1, with terms in powers of (x-1). This yields (x-1) - $(1/2)(x-1)^2$ + $(1/3)(x-1)^3$. A tempting distractor like choice B fails by including an unnecessary constant term of 1, which contradicts f(1)=0. Always compute successive derivatives and divide by the appropriate factorial to build accurate Taylor polynomials for any function.

We need to find the fifth-degree Taylor polynomial for f(x) = arctan x centered at x = 0. The derivative f'(x) = 1/(1+x²) can be expanded as a geometric series 1 - x² + x⁴ - x⁶ + ..., and integrating term by term gives arctan x = x - x³/3 + x⁵/5 - x⁷/7 + .... Therefore, the fifth-degree Taylor polynomial is P₅(x) = x - x³/3 + x⁵/5. Choice B incorrectly has all positive signs, missing the alternating pattern that comes from the geometric series expansion of 1/(1+x²). The key strategy is recognizing that arctan x integrates from the geometric series of its derivative.

This problem requires finding the degree-4 Taylor polynomial for f(x) = 1/(1-x) centered at x = 0. We can recognize this as the sum of a geometric series: 1/(1-x) = 1 + x + x² + x³ + x⁴ + ... for |x| < 1. Therefore, the degree-4 Taylor polynomial is simply the first five terms: P₄(x) = 1 + x + x² + x³ + x⁴. We can verify by derivatives: f(0) = 1, f'(x) = 1/(1-x)² gives f'(0) = 1, f''(x) = 2/(1-x)³ gives f''(0) = 2, and so on, yielding coefficients 1, 1, 2/2!, 6/3!, 24/4! = 1, 1, 1, 1, 1. Choice B incorrectly alternates signs, confusing this with 1/(1+x). Remember that 1/(1-x) generates all positive coefficients equal to 1, while 1/(1+x) alternates signs.

This problem asks for the degree-4 Taylor polynomial of $g(x) = \ln(1+x)$ centered at $x = 0$. The Taylor polynomial requires computing derivatives at $x = 0$: $g(0) = \ln(1) = 0$, $g'(x) = \frac{1}{1+x}$ gives $g'(0) = 1$, $g''(x) = -\frac{1}{(1+x)^2}$ gives $g''(0) = -1$, $g'''(x) = \frac{2}{(1+x)^3}$ gives $g'''(0) = 2$, and $g^{(4)}(x) = -\frac{6}{(1+x)^4}$ gives $g^{(4)}(0) = -6$. Applying the Taylor formula: $P_4(x) = 0 + 1 \cdot x + (-1) \frac{x^2}{2!} + 2 \frac{x^3}{3!} + (-6) \frac{x^4}{4!} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$. Choice B incorrectly has all positive terms, missing the alternating sign pattern that comes from the derivatives of $\ln(1+x)$. The key insight is that $\ln(1+x)$ generates an alternating series with coefficients $(-1)^{n+1}/n$ for the $x^n$ term.

This problem requires finding the degree-3 Taylor polynomial for f(x) = e^(2x) centered at x = 0. The Taylor polynomial formula uses P₃(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3!. Computing derivatives: f(x) = e^(2x) gives f(0) = 1, f'(x) = 2e^(2x) gives f'(0) = 2, f''(x) = 4e^(2x) gives f''(0) = 4, and f'''(x) = 8e^(2x) gives f'''(0) = 8. Substituting these values yields P₃(x) = 1 + 2x + 4x²/2 + 8x³/6 = 1 + 2x + 2x² + (4/3)x³. Choice B incorrectly uses 4x² instead of 2x², likely forgetting to divide by 2!. Remember that the Taylor polynomial coefficient for $x^n$ is f^(n)(a)/n!, where you must divide the nth derivative by n factorial.

Constructing Taylor polynomials involves approximating a function using its derivatives at a specific point, here x=0 for sin x up to degree 5. The constant term is f(0)=sin(0)=0, so no constant. The first derivative f'(x)=cos x, f'(0)=1, giving coefficient 1 for x; second f''(x)=-sin x, f''(0)=0; third f'''(x)=-cos x, f'''(0)=-1, coefficient -1/3! for x³; fourth $f^{(4)}$(x)=sin x=0 at 0; fifth $f^{(5)}$(x)=cos x=1 at 0, coefficient 1/5! for x⁵. This pattern shows odd powers with alternating signs divided by factorials. A tempting distractor like choice D omits factorials, using 1/3 and 1/5 instead of 1/3! and 1/5!, which underestimates the denominator and fails to match the series. To construct Taylor polynomials in general, evaluate the function and its derivatives at the center point and use the formula P_n(x) = sum_${k=0}^n$ $f^{(k)}$(a) / k! $(x-a)^k$.

This problem asks for the degree-4 Taylor polynomial of $f(x) = e^x$ centered at $x = 1$. For exponential functions, all derivatives equal $e^x$, so $f(1) = f'(1) = f''(1) = f'''(1) = f^{(4)}(1) = e$. The Taylor polynomial centered at $x = 1$ uses powers of $(x-1)$: $P_4(x) = e + e(x-1) + \frac{e(x-1)^2}{2!} + \frac{e(x-1)^3}{3!} + \frac{e(x-1)^4}{4!}$. Factoring out e: $P_4(x) = e[1 + (x-1) + \frac{(x-1)^2}{2} + \frac{(x-1)^3}{6} + \frac{(x-1)^4}{24}]$. Choice E incorrectly omits the factor of e, forgetting that all derivatives of $e^x$ evaluated at $x = 1$ equal e, not 1. Remember that when the center is not zero, you must evaluate the function and its derivatives at the center point.