- Match the coefficient of \((x-1)^2\) with the Taylor coefficient formula. - Remember the factor of \(2!\).

1. In a Taylor polynomial centered at \(1\), the coefficient of \((x-1)^2\) is \(\frac{f''(1)}{2!}\). 2. Therefore, \(\frac{f''(1)}{2}=3\), so \(f''(1)=6\).

\(f''(1)=6\).

- First find the displacement from the center \(3\). - Substitute that displacement directly into the given polynomial.

1. Substitute \(x=3.1\), so \(x-3=0.1\). 2. Then \(P_2(3.1)=5+2(0.1)-(0.1)^2=5.19\).

\(f(3.1)\approx5.19\).

- Identify the largest exponent whose coefficient is not zero. - Count actual nonzero terms separately from the exponents that are missing. - Compare what each quantity is measuring rather than expecting them to count the same thing.

1. The highest power with a nonzero coefficient is \(x^6\), so the degree is \(6\). 2. The polynomial has three nonzero terms: the constant term, the quadratic term, and the sixth-degree term. 3. Degree records the highest exponent with a nonzero coefficient. It does not count how many lower powers actually appear, so missing powers do not reduce the degree unless the highest nonzero power changes.

a) \(6\). b) \(3\) nonzero terms. c) Degree is determined by the highest nonzero power, not by the number of nonzero terms.

- Use the factorial form of a Taylor polynomial. - Which derivative supplies the coefficient of the squared term?

1. A degree-2 Taylor polynomial has quadratic coefficient \(\frac{f''(2)}{2!}\). 2. This is \(\frac{6}{2}=3\), so choice B is correct.

B: \(4-3(x-2)+3(x-2)^2\). The quadratic coefficient is \(\frac{f''(2)}{2!}=3\).

- Follow the vertical guide to compare all three curve heights at the same input. - A model can be below the function and still be close. Compare distances, not just which curve is higher.

1. At \(x=0.5\), the curve for \(q\) lies closer vertically to the curve for \(f\) than the line for \(t\). 2. Thus the quadratic Taylor model \(q\) is closer at the marked input. Its curve lies below \(f\).

The quadratic model \(q\); it is below \(f\) at \(x=0.5\).

- What information do you need to build a quadratic Taylor polynomial at a specified center? - The given first derivative tells you the coefficient of the linear term. - Use the displacement from the center, not the full new \(x\)-coordinate, in the polynomial.

1. The degree-2 Taylor polynomial centered at \(a=\sqrt{2}\) is \(P_2(x)=y(a)+y'(a)(x-a)+\frac{y''(a)}{2}(x-a)^2\). 2. Substituting the given values gives \(P_2(x)=\sqrt{2}-\frac{3\sqrt{2}}{16}(x-\sqrt{2})^2\). 3. At \(x=\sqrt{2}+\frac{1}{5}\), the estimate is \(P_2(x)=\sqrt{2}-\frac{3\sqrt{2}}{16}\cdot\frac{1}{25}=\frac{397\sqrt{2}}{400}\).

\(P_2(x)=\sqrt{2}-\frac{3\sqrt{2}}{16}(x-\sqrt{2})^2\); estimated value \(\frac{397\sqrt{2}}{400}\).

- Write the Taylor formula before substituting derivative values. - Divide the second-derivative value by \(2!\). - Evaluate using the displacement from the center, not the full input value.

1. The degree-\(2\) Taylor polynomial centered at \(2\) is \( P_2(x)=f(2)+f'(2)(x-2)+\frac{f''(2)}{2!}(x-2)^2\). 2. Substituting the given data gives \( P_2(x)=3-4(x-2)+3(x-2)^2\). 3. At \(x=2.05\), the change from the center is \(0.05\). Therefore, \( P_2(2.05)=3-4(0.05)+3(0.05)^2=2.8075\).

a) \( P_2(x)=3-4(x-2)+3(x-2)^2\). b) \(f(2.05)\approx2.8075\).

- Organize the polynomial by degree and ask what information determines each coefficient. - Compare the proposed coefficients with the standard Maclaurin-coefficient pattern. - Check the higher-degree terms more carefully than the constant and linear terms.

1. The degree-\(3\) Maclaurin formula is \(P_3(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3\). 2. Substitution gives \(P_3(x)=2+3x-2x^2+2x^3\). 3. Eemeli used derivative values as coefficients without dividing the second- and third-derivative values by their factorials.

a) \(P_3(x)=2+3x-2x^2+2x^3\). b) Eemeli omitted the factors \(2!\) and \(3!\) in the Taylor coefficients.

- Replace \(x\) with the center plus the displacement \(h\). - Expand before translating back to powers of \(x-1\). - Compare the degree of the function with the degree of the Taylor polynomial.

1. Since \(x=1+h\), substitute into the function: \( f(1+h)=(1+h)^3-2(1+h)\). 2. Expanding and combining terms gives \( f(1+h)=-1+h+3h^2+h^3\). 3. Replacing \(h\) by \(x-1\), the Taylor polynomial is \( P_3(x)=-1+(x-1)+3(x-1)^2+(x-1)^3\). 4. The original function is already a polynomial of degree \(3\), so its Taylor polynomial through degree \(3\) contains every term and is exact everywhere.

a) \( f(1+h)=-1+h+3h^2+h^3\). b) \( P_3(x)=-1+(x-1)+3(x-1)^2+(x-1)^3\). c) A degree-\(3\) polynomial is reproduced exactly by its third-degree Taylor polynomial.

- Check the derivatives used to form the Taylor polynomial. - Compare the error curve with the x-axis at the marked inputs. - Think about the derivatives of a difference of two functions that agree through second order.

1. \(f(0)=f'(0)=f''(0)=1\), so \(P_2(x)=1+x+\frac{x^2}{2}\). 2. The curve is below the x-axis at \(x=-0.5\), giving \(f(-0.5)<p(-0.5)\), and above it at \(x=0.5\), giving \(f(0.5)>p(0.5)\). 3. Matching value and first two derivatives gives \(d(0)=d'(0)=d''(0)=0\), explaining the flat contact near zero.

a) \(p(x)=1+x+\frac{x^2}{2}\) matches the value and first two derivatives at \(0\). b) \(d(-0.5)<0\) and \(d(0.5)>0\); the model is above \(f\) at the negative input and below it at the positive input. c) \(d(0)=d'(0)=d''(0)=0\).

- Relate the Taylor polynomial for an antiderivative to the polynomial for its derivative, while also using the antiderivative's value at the center. - After constructing the polynomial, connect its fourth-power coefficient back to a derivative at the center. - Evaluate the resulting Taylor polynomial only after its coefficients have been justified.

1. Because \(F'(x)=\ln(1+x)\), the derivative of the Maclaurin polynomial for \(F\) must match \(x-\frac{x^2}{2}+\frac{x^3}{3}\) through degree \(3\). Since \(F(0)=0\), termwise integration gives \(Q_4(x)=\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}\). 2. In a Maclaurin polynomial, the coefficient of \(x^4\) is \(\frac{F^{(4)}(0)}{4!}\). Therefore, \(\frac{F^{(4)}(0)}{24}=\frac{1}{12}\), so \(F^{(4)}(0)=2\). 3. At \(x=0.2\), \(Q_4(0.2)=\frac{(0.2)^2}{2}-\frac{(0.2)^3}{6}+\frac{(0.2)^4}{12}=0.0188\).

a) \(Q_4(x)=\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}\). b) \(F^{(4)}(0)=2\). c) \(F(0.2)\approx0.0188\).

- A Maclaurin coefficient is a derivative value divided by the appropriate factorial. - The derivatives of \(f'\) at the center are the next-order derivatives of \(f\). - Build the Taylor polynomial for \(f'\) from those center data before evaluating it.

1. Matching coefficients with \(\frac{f^{(k)}(0)}{k!}x^k\) gives \(f'(0)=1\), \(f''(0)=2!(-3)=-6\), \(f'''(0)=3!\left(\frac12\right)=3\), and \(f^{(4)}(0)=4!(1)=24\). 2. The degree-\(3\) Maclaurin polynomial for \(f'\) is \(1-6x+\frac{3}{2!}x^2+\frac{24}{3!}x^3\), so \(1-6x+\frac32x^2+4x^3\). 3. At \(x=0.2\), the approximation is \(1-6(0.2)+\frac32(0.2)^2+4(0.2)^3=-0.108\).

a) \(f'(0)=1\), \(f''(0)=-6\), \(f'''(0)=3\), \(f^{(4)}(0)=24\). b) \(1-6x+\frac32x^2+4x^3\). c) \(f'(0.2)\approx-0.108\).

- First use the polynomial value to determine the unknown quadratic coefficient. - Once the coefficient is known, compare it with the standard Taylor coefficient for the second-order term. - The constant and linear Taylor coefficients encode the function value and first derivative at the center.

1. Substituting \(x=1.2\) gives \(2.68=2+3(0.2)+c(0.2)^2\). Thus \(2.68=2.6+0.04c\), so \(c=2\). 2. For a Taylor polynomial centered at \(1\), the quadratic coefficient is \(\frac{f''(1)}{2!}\). Hence \(2=\frac{f''(1)}{2}\), so \(f''(1)=4\). 3. The constant and linear coefficients give \(f(1)=2\) and \(f'(1)=3\).

a) \(c=2\). b) \(f''(1)=4\). c) \(f(1)=2\) and \(f'(1)=3\).

- Use the symmetry information before writing the polynomial. - Divide each even derivative by its factorial. - Evaluate the quadratic and quartic contributions separately.

1. Because \(f\) is even, every odd derivative at \(0\) is \(0\). 2. The polynomial is \( P_4(x)=2+\frac{-6}{2!}x^2+\frac{24}{4!}x^4=2-3x^2+x^4\). 3. The odd-power terms are absent because an even function has a Maclaurin series containing only even powers. 4. At \(x=0.3\), \( P_4(0.3)=2-3(0.3)^2+(0.3)^4=1.7381\).

a) \( P_4(x)=2-3x^2+x^4\). b) Even symmetry forces the odd derivatives at \(0\) to equal \(0\). c) \(f(0.3)\approx1.7381\).

- Substitute the entire inner expression into every occurrence of the polynomial variable. - Apply the power to the factor \(3\) as well as to \(x\). - Simplify the polynomial before evaluating it.

1. Replace the input \(u\) in the polynomial by \(3x\): \( Q_3(x)=1+2(3x)-(3x)^2+(3x)^3\). 2. Simplifying gives \( Q_3(x)=1+6x-9x^2+27x^3\). 3. At \(x=0.1\), \( Q_3(0.1)=1+0.6-0.09+0.027=1.537\).

a) \( Q_3(x)=1+6x-9x^2+27x^3\). b) \(g(0.1)\approx1.537\).

- Use what a Maclaurin polynomial matches at the center before reading the comparison away from the center. - At the marked input, compare vertical separation from \(f\); the magnified panel is intended to distinguish \(q\), \(r\), and \(f\). - For the explanation, connect polynomial degree with how many derivatives are matched at the center.

1. Since these are Maclaurin polynomials for \(\sin x\), all four functions have value \(0\) at the center and slope \(1\) there. 2. At \(x=2\), panel a) shows that \(p\) is far above \(f\). Panel b) shows that \(q\) lies below \(f\) while \(r\) lies just above \(f\), and the vertical separation between \(r\) and \(f\) is smaller than the separation between \(q\) and \(f\). 3. Therefore the absolute-error ranking from smallest to largest is \(r\), \(q\), \(p\). At \(x=2\), \(p\) and \(r\) lie above \(f\), while \(q\) lies below \(f\). 4. Increasing the Taylor degree matches more derivatives of \(\sin x\) at \(0\). Consequently, the first possible nonzero local error term occurs at a higher power of \(x\), which is small near the center.

a) Shared value \(0\) and shared slope \(1\). b) From smallest to largest absolute error at \(x=2\): \(r\), \(q\), \(p\). The curves \(p\) and \(r\) are above \(f\), and \(q\) is below \(f\). c) Higher-degree Taylor polynomials match more derivatives of \(\sin x\) at \(0\), pushing the first possible local error to higher degree.

- A degree-\(1\) Taylor polynomial uses the function value and first derivative at its own center. - Evaluate each local model at the same target input before comparing their center distances. - Ask what controls Taylor error besides the size of \(|x-a|\).

1. At center \(0\), \(P_1(x)=f(0)+f'(0)x=1+3x\). 2. At center \(1\), \(Q_1(x)=f(1)+f'(1)(x-1)=2-(x-1)=3-x\). 3. At \(x=0.8\), \(P_1(0.8)=3.4\) and \(Q_1(0.8)=2.2\). 4. The displacements are \(|0.8-0|=0.8\) and \(|0.8-1|=0.2\). The second approximation is based closer to its center, but distance alone does not determine which approximation is more accurate. Accuracy also depends on higher derivatives and the corresponding Taylor remainders, none of which are given.

a) \(P_1(x)=1+3x\) and \(Q_1(x)=3-x\). b) \(P_1(0.8)=3.4\) and \(Q_1(0.8)=2.2\). c) The center \(1\) is closer to \(0.8\), but that fact alone does not determine which approximation is more accurate for \(f(0.8)\) without additional remainder information.

- Relate the coefficient in row \(k\) to the \(k\)th derivative at the center. - Remember that the factorial is part of the Taylor coefficient. - Treat the constant and linear rows as the first two cases of the same coefficient rule.

1. In a Taylor polynomial centered at \(2\), \(c_k=\frac{f^{(k)}(2)}{k!}\). 2. Therefore, \(f(2)=c_0=-1\) and \(f'(2)=c_1=4\). 3. Since \(c_2=3\), \(f''(2)=2!\cdot3=6\). 4. Since \(c_3=-2\), \(f'''(2)=3!\cdot(-2)=-12\).

\(f(2)=-1\), \(f'(2)=4\), \(f''(2)=6\), and \(f'''(2)=-12\).

- Subtract the polynomial from the function before substituting either input. - Compare the error magnitudes rather than the input distances alone.

1. The error is \(|f(x)-P_2(x)|=|x^3(x-0.3)|\). 2. At \(x=0.1\), this is \(0.0002\). At \(x=0.3\), the error is \(0\). 3. The farther input \(0.3\) gives the smaller actual error. Distance from the center alone does not determine actual error.

Errors: \(0.0002\) at \(0.1\), and \(0\) at \(0.3\). No—the farther input can have the smaller error.

- Match the square term to the second derivative at the center. - Pay attention to the factorial in the Taylor coefficient formula.

1. In the Taylor formula, the quadratic coefficient is \(\frac{f''(-1)}{2!}\). 2. Thus \(k=\frac{-8}{2}=-4\).

\(k=-4\), since \(k=\frac{f''(-1)}{2!}\).

- Interpret position, velocity, and acceleration as a function and its first two derivatives. - Form the Taylor polynomial using time measured from the reference instant. - Ask what would be needed to certify the model's error.

1. The Taylor model is \(P_2(t)=120+22(t-4)+\frac{-4}{2}(t-4)^2\), measured in feet. 2. At \(t=4.5\), the displacement from the center is \(0.5\), giving \(120+11-0.5=130.5\,\text{ft}\). 3. No bound on higher derivatives or remainder is supplied, so this estimate has no guaranteed error size from the given data alone.

Approximately \(130.5\,\text{ft}\) from the quadratic Taylor model. No error guarantee is available.

- Compute the first three derivatives of the exponential function at \(0\). - Write the corresponding derivatives of a general cubic polynomial. - Evaluate the polynomial and its derivatives at \(0\). - Match corresponding values to determine the coefficients.

1. Compute the derivatives of \(f\): \(f'(x)=-2e^{-2x}\), \(f''(x)=4e^{-2x}\), and \(f'''(x)=-8e^{-2x}\). At \(x=0\), the values are \(f(0)=1\), \(f'(0)=-2\), \(f''(0)=4\), and \(f'''(0)=-8\). 2. For the cubic polynomial, \(p(0)=a_0\), \(p'(0)=a_1\), \(p''(0)=2a_2\), and \(p'''(0)=6a_3\). Matching the values gives \(a_0=1\), \(a_1=-2\), \(2a_2=4\), and \(6a_3=-8\). Therefore, \(a_2=2\) and \(a_3=-\frac{4}{3}\), so \(p(x)=-\frac{4}{3}x^3+2x^2-2x+1\).

\(p(x)=-\frac{4}{3}x^3+2x^2-2x+1\)

- Begin with the product rule for the first derivative, then differentiate that identity once more for the second derivative. - After finding the derivative values of \(h\) at the center, place them into the factorial form of a Maclaurin polynomial. - Evaluate the Taylor polynomial only after its coefficients have been established from the derivative data.

1. The function value is \(h(0)=f(0)g(0)=1\). 2. By the product rule, \(h'(0)=f'(0)g(0)+f(0)g'(0)=1-1=0\). 3. Differentiating the product rule once more gives \(h''(0)=f''(0)g(0)+2f'(0)g'(0)+f(0)g''(0)\). Substitution gives \(h''(0)=1-2+2=1\). 4. Therefore the degree-\(2\) Maclaurin polynomial is \(H_2(x)=h(0)+h'(0)x+\frac{h''(0)}{2!}x^2=1+\frac{x^2}{2}\). 5. Thus \(h(0.1)\approx H_2(0.1)=1+\frac{(0.1)^2}{2}=1.005\).

a) \(h(0)=1\), \(h'(0)=0\), and \(h''(0)=1\), using \(h'=f'g+fg'\) and \(h''=f''g+2f'g'+fg''\). b) \(H_2(x)=1+\frac{x^2}{2}\). c) \(h(0.1)\approx1.005\).

- Rewrite the input as the center plus a small displacement. - Use the angle-addition identity before substituting Maclaurin terms. - Keep only powers through degree \(3\) in the displacement.

1. Let \(h=x-\frac{\pi}{4}\). The angle-addition identity gives \( \sin\left(\frac{\pi}{4}+h\right)=\frac{\sqrt2}{2}(\cos h+\sin h)\). 2. Through degree \(3\), \( \cos h=1-\frac{h^2}{2}+\cdots\) and \( \sin h=h-\frac{h^3}{6}+\cdots\). 3. Therefore, \( P_3(x)=\frac{\sqrt2}{2}\left(1+h-\frac{h^2}{2}-\frac{h^3}{6}\right)\), where \(h=x-\frac{\pi}{4}\). 4. At \(h=0.1\), \( P_3\approx\frac{\sqrt2}{2}\left(1.1-0.005-0.000166667\right)\approx0.774164\).

\( P_3(x)=\frac{\sqrt2}{2}\left(1+h-\frac{h^2}{2}-\frac{h^3}{6}\right)\), where \(h=x-\frac{\pi}{4}\). \( \sin\left(\frac{\pi}{4}+0.1\right)\approx0.774164\).

- Express the polynomial equation in terms of \(x-3\). - Distinguish closeness to the expansion center from known approximation error. - Ask what information you have about the original function away from the center.

1. Set \(h=x-3\). Then \(h^2-2h+0.75=0=(h-0.5)(h-1.5)\). 2. Thus the polynomial solutions are \(x=3.5\) and \(x=4.5\). 3. \(3.5\) lies nearer the center \(3\). That fact alone does not establish the actual approximation error at either input. The Taylor information is local and comes with no remainder guarantee here. 4. Neither solution is known to solve the equation for \(f\). The original function may behave differently away from the center.

a) \(x=3.5\) and \(x=4.5\). b) \(3.5\) is nearer the center, but this does not prove that it is more accurate. c) Neither is guaranteed to solve \(f(x)=1.75\).

- Rewrite the absolute-value function separately for positive and negative inputs. - Check derivatives at \(0\) using limits rather than only formulas valid away from \(0\). - A Taylor polynomial requires every derivative through its stated degree.

1. Clearly, \(f(0)=0\). 2. For \(x>0\), \(f(x)=x^3\); for \(x<0\), \(f(x)=-x^3\). Thus, for \(x\neq0\), \( f'(x)=3x|x|\). Also, \( f'(0)=\lim_{h\to0}\frac{|h|^3}{h}=0\). 3. Using the limit definition again, \( f''(0)=\lim_{h\to0}\frac{f'(h)-f'(0)}{h}=\lim_{h\to0}3|h|=0\). For \(x\neq0\), \(f''(x)=6|x|\). 4. The degree-\(2\) Maclaurin polynomial is \(P_2(x)=0\). 5. The one-sided derivatives of \(f''(x)=6|x|\) at \(0\) are \(-6\) and \(6\), so \(f'''(0)\) does not exist. Therefore, a degree-\(3\) Maclaurin polynomial is not defined.

a) \(f(0)=f'(0)=f''(0)=0\). b) \(P_2(x)=0\). c) \(f'''(0)\) does not exist because its one-sided values disagree.

- Read the Taylor coefficients with their factorials. - Separate a derivative value at one point from information about a whole interval. - Distinguish features of the polynomial model from guaranteed features of the original function.

1. The coefficients give \(f(2)=5\), \(f'(2)=-2\), and \(f''(2)=2!\cdot4=8\). 2. At \(x=2\), the tangent slope is negative and the second derivative is positive. Since \(f'\) is differentiable at \(2\), it is continuous there, so \(f\) decreases on some small interval around \(2\). Neither sign determines the behavior throughout \([2,3]\), and \(f''(2)>0\) alone does not establish concavity on an interval. 3. With \(h=x-2\), the model is \(5-2h+4h^2\), whose derivative is \(-2+8h\). It is zero at \(h=\frac{1}{4}\), so the modeled minimum is at \(x=2.25\) with value \(4.75\). 4. A model's minimum need not be an actual minimum of \(f\); no error or broader shape information is given.

a) \(f(2)=5\), \(f'(2)=-2\), \(f''(2)=8\). b) The slope is negative and the second derivative positive at \(2\). Local decrease follows, but decrease and concavity throughout \([2,3]\) are not established. c) The model minimum is \(4.75\) at \(x=2.25\). d) No conclusion about an actual minimum of \(f\) follows.

- Give the reciprocal polynomial one unknown coefficient for each power through \(x^2\). - In the product, the constant coefficient must be \(1\), while the coefficients of \(x\) and \(x^2\) must be \(0\). - Solve the coefficient equations in increasing order of degree.

1. Multiply \( (2+x-x^2)(a+bx+cx^2)\) and match coefficients through degree \(2\). 2. The constant equation is \(2a=1\), so \(a=1/2\). 3. The coefficient of \(x\) gives \(2b+a=0\), so \(b=-1/4\). 4. The coefficient of \(x^2\) gives \(2c+b-a=0\). Substituting \(a=1/2\) and \(b=-1/4\) gives \(c=3/8\). 5. Therefore, \( Q_2(x)=\frac12-\frac{x}{4}+\frac{3x^2}{8}\). 6. Multiplication gives \( P_2(x)Q_2(x)=1+\frac{5}{8}x^3-\frac{3}{8}x^4\), so the product equals \(1\) through degree \(2\), as required.

\( Q_2(x)=\frac12-\frac{x}{4}+\frac{3x^2}{8}\).

- Treat \(x+x^2\) as the input to the exponential polynomial. - When expanding powers of the input, discard terms above degree \(3\). - Combine like powers only after all three polynomial terms have been substituted.

1. Let \(u=x+x^2\). Through degree \(3\), \( e^u=1+u+\frac{u^2}{2}+\frac{u^3}{6}+\) terms of degree greater than \(3\). 2. Compute only the powers needed through degree \(3\): \( u^2=x^2+2x^3+\) higher-degree terms, and \( u^3=x^3+\) higher-degree terms. 3. Substitution gives \( P_3(x)=1+(x+x^2)+\frac{x^2+2x^3}{2}+\frac{x^3}{6}=1+x+\frac{3x^2}{2}+\frac{7x^3}{6}\). 4. Therefore, \( f(0.1)\approx1+0.1+\frac{3(0.1)^2}{2}+\frac{7(0.1)^3}{6}\approx1.116166667\).

\( P_3(x)=1+x+\frac{3x^2}{2}+\frac{7x^3}{6}\). \( f(0.1)\approx1.116166667\).

- Use displacements \(0.1\) and \(-0.1\) from the center. - Compare how even and odd powers behave when the sign of the displacement changes. - In the difference quotient, subtract before dividing by the full distance between the inputs.

1. At \(x=1.1\), the displacement is \(0.1\), so \( P_3(1.1)=5+0.2-0.03+0.004=5.174\). 2. At \(x=0.9\), the displacement is \(-0.1\), so \( P_3(0.9)=5-0.2-0.03-0.004=4.766\). 3. Their average is \( \frac{5.174+4.766}{2}=4.970\). The odd-power terms cancel in this average. 4. The symmetric difference quotient is \( \frac{5.174-4.766}{0.2}=2.04\). The constant and even-power terms cancel in the numerator of this quotient.

a) \(f(1.1)\approx5.174\) and \(f(0.9)\approx4.766\). b) Average \(4.970\); odd-power terms cancel. c) Symmetric difference quotient \(2.04\); constant and even-power terms cancel.

- Replace \(x\) with \(1+(x-1)\) before expanding. - Collect powers of the displacement only after expanding every term. - Distinguish evaluating a polynomial from obtaining a guaranteed approximation to the original function.

1. Let \(h=x-1\), so \(x=1+h\). Substitute into the polynomial: \( P_4(1+h)=2-(1+h)+\frac32(1+h)^2-\frac12(1+h)^3+\frac14(1+h)^4\). 2. Expanding and collecting powers of \(h\) gives \( P_4(1+h)=\frac94+\frac32h+\frac32h^2+\frac12h^3+\frac14h^4\). 3. Therefore, \( P_4(x)=\frac94+\frac32(x-1)+\frac32(x-1)^2+\frac12(x-1)^3+\frac14(x-1)^4\). 4. At \(x=1.08\), the displacement is \(h=0.08\), so \( P_4(1.08)=2.37986624\). 5. The original Maclaurin polynomial matches derivatives of \(f\) at \(0\). Rewriting the same polynomial around \(1\) neither proves derivative matching there nor supplies an error bound at \(1.08\). Thus the number in part b) is only a polynomial value, not a guaranteed approximation to \(f(1.08)\).

a) \( P_4(x)=\frac94+\frac32(x-1)+\frac32(x-1)^2+\frac12(x-1)^3+\frac14(x-1)^4\). b) \( P_4(1.08)=2.37986624\). c) Rewriting does not transfer Taylor matching to the new center and provides no remainder bound for \(f(1.08)\).

- Apply factorial scaling when converting coefficients to derivatives. - The lowest-degree nonzero term controls the local behavior near the center. - Comparing opposite inputs cancels every even-power contribution. - For part d), substitute \(x=0.1\) only after identifying which polynomial terms are nonzero.

1. Comparing with Taylor form gives \(f(0)=5\), \(f'(0)=f''(0)=f'''(0)=0\), \(f^{(4)}(0)=4!(-3)=-72\), and \(f^{(5)}(0)=5!(2)=240\). 2. Near \(0\), the first nonconstant term is \(-3x^4\). It is negative on both sides of \(0\) and has lower degree than the later term \(2x^5\). Therefore, \(f(x)<f(0)\) for all sufficiently small nonzero \(x\), so \(x=0\) is a strict local maximum. 3. The even fourth-degree terms cancel when opposite inputs are compared, while the fifth-degree terms give \( P_5(x)-P_5(-x)=4x^5\). Thus, for sufficiently small positive \(x\), \(f(x)>f(-x)\). 4. The approximation is \( f(0.1)\approx5-3(0.1)^4+2(0.1)^5=4.99972\).

a) \(f(0)=5\), \(f'(0)=f''(0)=f'''(0)=0\), \(f^{(4)}(0)=-72\), and \(f^{(5)}(0)=240\). b) \(x=0\) is a strict local maximum. c) For small \(x>0\), \(f(x)>f(-x)\). d) \(f(0.1)\approx4.99972\).

- Connect the missing quadratic coefficient to the second derivative at the center. - Determine the center before substituting a shifted variable. - Remember that the coefficient of \((x-a)^n\) equals \(f^{(n)}(a)/n!\). - For part d), use the centered form and the displacement from the recovered center.

1. A missing quadratic term means the second derivative of the polynomial is \(0\) at the center. 2. Since \( P''(x)=12x-24\), solving \(P''(a)=0\) gives \(a=2\). 3. Let \(h=x-2\). Substituting \(x=2+h\) and collecting powers gives \( P(x)=3+(x-2)+2(x-2)^3\). 4. Comparing with Taylor form gives \(f(2)=3\), \(f'(2)=1\), \(f''(2)=0\), and \( \frac{f'''(2)}{3!}=2\), so \(f'''(2)=12\). 5. At \(x=2.1\), the displacement is \(0.1\). Therefore, \( f(2.1)\approx3+0.1+2(0.1)^3=3.102\).

a) \(a=2\). b) \( P(x)=3+(x-2)+2(x-2)^3\). c) \(f(2)=3\), \(f'(2)=1\), \(f''(2)=0\), and \(f'''(2)=12\). d) \(f(2.1)\approx3.102\).

- Differences of values at \(x\) and \(-x\) isolate odd powers. - Averages of values at opposite inputs isolate even powers. - Apply factorial scaling only after all polynomial coefficients are known.

1. Write \( P_4(x)=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4\). Since \(P_4(0)=3\), \(c_0=3\). 2. Opposite-input differences isolate the odd coefficients: \( \frac{P_4(1)-P_4(-1)}{2}=c_1+c_3=0\), and \( \frac{P_4(2)-P_4(-2)}{4}=c_1+4c_3=-3\). Thus \(c_3=-1\) and \(c_1=1\). 3. Opposite-input averages isolate the even coefficients: \(c_2+c_4=3\) and \(c_2+4c_4=6\). Hence \(c_4=1\) and \(c_2=2\). 4. Therefore, \( P_4(x)=3+x+2x^2-x^3+x^4\). 5. Comparing with Taylor form gives \(f(0)=3\), \(f'(0)=1\), \(f''(0)=4\), \(f'''(0)=-6\), and \(f^{(4)}(0)=24\). 6. Finally, \( f(0.3)\approx P_4(0.3)=3.4611\).

a) \( P_4(x)=3+x+2x^2-x^3+x^4\). b) \(f(0)=3\), \(f'(0)=1\), \(f''(0)=4\), \(f'''(0)=-6\), and \(f^{(4)}(0)=24\). c) \(f(0.3)\approx3.4611\).

- Which derivatives are fixed by knowing a polynomial through degree 3? - When squaring, which powers can involve the first unknown coefficient? - Consider what changes when the original function has zero constant term.

1. The degree-3 polynomial determines derivatives of \(f\) through order 3 but not its fourth derivative. Derivatives of \(f^2\) through order 3 are determined, so degree 3 is guaranteed. 2. Expanding and keeping terms through degree 3 gives \([P_3(x)]^2=1+2x+5x^2+2x^3+\cdots\). Hence \(P_3^{f^2}(x)=1+2x+5x^2+2x^3\). 3. Each \(f_c\) shares derivatives with \(P_3\) through third order. The degree-4 coefficient of \(f_c^2\) is \(2+2c\), which depends on \(c\). 4. In the new case, \(f\) has no constant term and its linear coefficient is nonzero. Its unknown fourth-degree coefficient cannot affect the square before degree 5. The smoothness assumption ensures the fourth-degree Maclaurin polynomial of \(f^2\) exists, so its coefficients through degree 4 are determined.

a) Degree \(3\). b) \(1+2x+5x^2+2x^3\). c) The coefficient of \(x^4\) in \(f_c^2\) is \(2+2c\). d) Degree \(4\).

- Expand only powers that can contribute through degree 4. - Match coefficients in order, starting with the constant term. - For the last part, compare the smallest possible power of \(x\) in each substituted term.

1. Through degree 4, \(u=x+x^2\), \(u^2=x^2+2x^3+x^4\), \(u^3=x^3+3x^4+\cdots\), and \(u^4=x^4+\cdots\). 2. Therefore \(P_4(x+x^2)=a_0+a_1x+(a_1+a_2)x^2+(2a_2+a_3)x^3+(a_2+3a_3+a_4)x^4+\cdots\). 3. Matching coefficients yields \(a_0=1\), \(a_1=2\), \(a_1+a_2=5\), \(2a_2+a_3=10\), and \(a_2+3a_3+a_4=20\). Thus \(a_2=3\), \(a_3=4\), and \(a_4=5\). 4. \(P_4(u)=1+2u+3u^2+4u^3+5u^4\). At \(u=0.1\), its value is \(1.2345\). 5. Since \(x+x^2\) starts with a degree-1 term, any \(u^k\) with \(k\geq5\) starts with a power of \(x\) of degree at least 5.

a) \(a_0=1\), \(a_1=2\), \(a_1+a_2=5\), \(2a_2+a_3=10\), \(a_2+3a_3+a_4=20\). b) \(P_4(u)=1+2u+3u^2+4u^3+5u^4\). c) \(f(0.1)\approx1.2345\). d) The substituted higher-degree terms begin at degree 5.

- Use the derivatives of sine and cosine at zero to build the requested polynomials. - Which nonzero powers remain after subtracting the cosine polynomial from 1? - Compare the first nonzero powers of the two Cartesian coordinate expressions near the pole.

1. The Maclaurin polynomials are \(P_2(\theta)=1-\frac{\theta^2}{2}\) for cosine and \(Q_3(\theta)=\theta-\frac{\theta^3}{6}\) for sine. 2. Near zero, \(r=1-\cos\theta\approx\frac{\theta^2}{2}\). 3. Thus the first nonzero terms are \(x(\theta)\approx\frac{\theta^2}{2}\) and \(y(\theta)\approx\frac{\theta^3}{2}\). 4. Differentiating those leading polynomial terms gives a slope estimate \(\frac{dy}{dx}\approx\frac{3\theta^2/2}{\theta}=\frac{3\theta}{2}\), which approaches zero as \(\theta\) approaches zero. The predicted limiting tangent line at the pole is \(y=0\).

\(P_2(\theta)=1-\frac{\theta^2}{2}\); \(Q_3(\theta)=\theta-\frac{\theta^3}{6}\). Leading terms: \(x(\theta)\approx\frac{\theta^2}{2}\), \(y(\theta)\approx\frac{\theta^3}{2}\). Predicted limiting tangent: \(y=0\).