Free math worksheets

- 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.

- 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.

- Match the value and first two derivatives of the exponential function at the center. - At each vertical guide, read whether the error curve is above or below the x-axis rather than calculating its value. - Translate the sign of \(f-p\) into an ordering of \(f\) and \(p\).

1. For \(f(x)=e^x\), \(f(0)=1\), \(f'(0)=1\), and \(f''(0)=1\). Therefore the degree-\(2\) Maclaurin polynomial is \(1+x+\frac{x^2}{2}=p(x)\). 2. At the guide \(x=-0.5\), the graph of \(d\) lies below the x-axis, so \(d(-0.5)<0\). Since \(d=f-p\), this means \(f(-0.5)<p(-0.5)\). 3. At the guide \(x=0.5\), the graph of \(d\) lies above the x-axis, so \(d(0.5)>0\). Therefore \(f(0.5)>p(0.5)\). 4. Because \(p\) matches the value, first derivative, and second derivative of \(f\) at \(0\), \(d(0)=d'(0)=d''(0)=0\). Hence the error has no constant, linear, or quadratic local term, which explains the flat contact with the x-axis near the center.

a) \(p(x)=1+x+\frac{x^2}{2}\) is the degree-\(2\) Maclaurin polynomial for \(e^x\). b) \(d(-0.5)<0\), so \(p(-0.5)>f(-0.5)\). Also, \(d(0.5)>0\), so \(f(0.5)>p(0.5)\). c) \(d(0)=d'(0)=d''(0)=0\), so the error has high-order contact with \(0\) at the center.

- 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) \(g(0.1)\approx1.537\).

- Substitute the inner expression into the standard cosine series before truncating. - Track what happens to the even powers of \(\sqrt{x}\). - For the numerical approximation, choose the value of \(x\) whose square root is \(0.5\).

1. The cosine series is \(\cos u=1-\frac{u^2}{2!}+\frac{u^4}{4!}-\frac{u^6}{6!}+\cdots\). 2. Substitute \(u=\sqrt{x}\). Each even power of \(\sqrt{x}\) becomes an integer power of \(x\): \(\cos(\sqrt{x})=1-\frac{x}{2!}+\frac{x^2}{4!}-\frac{x^3}{6!}+\cdots\). 3. Therefore, \(P_3(x)=1-\frac{x}{2}+\frac{x^2}{24}-\frac{x^3}{720}\). 4. To approximate \(\cos(0.5)\), use \(x=(0.5)^2=0.25\): \(P_3(0.25)=1-\frac{0.25}{2}+\frac{0.25^2}{24}-\frac{0.25^3}{720}=\frac{40439}{46080}\approx0.877582465\).

a) \(P_3(x)=1-\frac{x}{2}+\frac{x^2}{24}-\frac{x^3}{720}\). b) \(\cos(0.5)\approx0.877582465\).

- 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 \(f'''(2)=-12\).

- Solve \(1+x=1.2\) before substituting. - Keep track of the alternating signs. - Compare the required value of \(x\) for \(\ln(4)\) with the given interval.

1. To approximate \(\ln(1.2)\), set \(1+x=1.2\), so \(x=0.2\). 2. The fourth-degree approximation is \(0.2-\frac{0.2^2}{2}+\frac{0.2^3}{3}-\frac{0.2^4}{4}=0.182266\ldots\). 3. Therefore, \(\ln(1.2)\approx0.1823\). 4. For \(\ln(4)\), direct substitution would require \(1+x=4\), or \(x=3\). Since \(3\) is outside the stated interval \((-1,1]\), the series representation does not apply there.

1. \(\ln(1.2)\approx0.1823\) 2. Direct substitution would require \(x=3\), which is outside \((-1,1]\).

- 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 \(h=x-\frac{\pi}{4}\). \( \sin\left(\frac{\pi}{4}+0.1\right)\approx0.774164\).

- Solve first in terms of the displacement \(x-3\). - After finding both model roots, recall where the data defining a centered Taylor polynomial are matched to the original function. - Keep separate the equation solved by the polynomial model and the equation involving the original function.

1. Let \(h=x-3\). Then \(1+2h-h^2=1.75\), so \(h^2-2h+0.75=0\). 2. Factoring gives \((h-0.5)(h-1.5)=0\). Thus \(h=0.5\) or \(h=1.5\), giving model solutions \(x=3.5\) and \(x=4.5\). 3. A Taylor polynomial centered at \(3\) is anchored by function and derivative data at \(x=3\). Of the two model solutions, \(x=3.5\) is closer to that center, so it is the better local estimate based on the available Taylor information. 4. The Taylor polynomial is not stated to equal \(f\) exactly away from the center. Therefore, neither model root is proved to satisfy \(f(x)=1.75\) exactly.

a) \(x=3.5\) and \(x=4.5\). b) \(x\approx3.5\) is the better estimate associated with the Taylor information at \(x=3\), because it is closer to the center where the polynomial matches \(f\)'s derivative data. c) Neither value is proved to be an exact solution of \(f(x)=1.75\) from the Taylor polynomial alone.

- 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 constant, linear, and quadratic coefficients using the factorial form of a Taylor polynomial. - Separate statements forced by derivative data at the center from predictions made by the polynomial away from the center. - Ask what additional information would be needed to transfer a feature of the model to the original function.

1. From the Taylor form, \(f(2)=5\), \(f'(2)=-2\), and \(\frac{f''(2)}{2!}=4\), so \(f''(2)=8\). 2. Since \(f'(2)<0\), the function is decreasing at the center. Since \(f''(2)>0\), it is concave up there. 3. Let \(h=x-2\). The model is \(5-2h+4h^2\). Its derivative with respect to \(h\) is \(-2+8h\), which is \(0\) at \(h=\frac14\). Therefore, the model has its minimum at \(x=2.25\). 4. The modeled minimum value is \(5-2(0.25)+4(0.25)^2=4.75\). 5. Part c establishes only a feature of the quadratic approximation. The Taylor polynomial is guaranteed to match specified derivative data at the center, but no error bound or exact agreement with \(f\) away from the center is given. Thus part c alone does not determine an actual minimum of \(f\).

a) \(f(2)=5\), \(f'(2)=-2\), \(f''(2)=8\). b) \(f\) is decreasing and concave up at \(x=2\). c) The quadratic model has its minimum at \(x=2.25\), with modeled value \(4.75\). d) Part c alone does not establish the location or value of an actual minimum of \(f\); it establishes only the minimum of the Taylor model.

- 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) \(f(0.3)\approx3.4611\).

- Separate what the given degree-3 polynomial determines from what remains unknown about the function. - Ask at which degree unknown higher-order behavior could first affect the square. - Use the family in part c) to test whether the next coefficient is forced or can vary. - In part d), pay attention to the lowest nonzero term when deciding when unknown behavior can first contribute.

1. A degree-\(3\) Maclaurin polynomial determines \(f(0)\), \(f'(0)\), \(f''(0)\), and \(f'''(0)\), but it gives no information about \(f^{(4)}(0)\). 2. The derivatives of \(f^2\) through order \(3\) depend only on those known derivatives of \(f\). Equivalently, multiplying the known coefficients and retaining powers through degree \(3\) gives \([P_3(x)]^2=1+2x+5x^2+2x^3+\text{higher-degree terms}\). Therefore, the guaranteed Maclaurin polynomial is \(1+2x+5x^2+2x^3\). 3. Every function \(f_c(x)=P_3(x)+cx^4\) has the same derivatives through order \(3\) at \(0\). The coefficient of \(x^4\) in \([P_3(x)]^2\) is \(2\), and the cross term \(2P_3(x)(cx^4)\) contributes \(2c\) because \(P_3(0)=1\). Thus the possible \(x^4\) coefficient is \(2+2c\), so it is not determined. 4. In the alternate case, the known polynomial begins with a nonzero linear term. The unknown fourth-degree behavior first combines with that linear term to affect degree \(5\). Therefore, the coefficients of \(f^2\) through degree \(4\) are guaranteed.

a) The degree-\(3\) Maclaurin polynomial is the highest one guaranteed. b) \(1+2x+5x^2+2x^3\). c) The coefficient of \(x^4\) in \(f_c(x)^2\) is \(2+2c\), so it is not fixed by the given polynomial. d) The Maclaurin polynomial for \(f^2\) is guaranteed through degree \(4\).

- Expand only the powers of \(x+x^2\) that can contribute through degree \(4\). - Match coefficients from lowest degree to highest; each equation introduces one new unknown. - For the last part, compare the lowest power of \(x\) in \((x+x^2)^k\) with the value of \(k\). - In part d), note that \((x+x^2)^k\) has lowest degree \(k\).

1. The powers needed through degree \(4\) are \(u=x+x^2\), \(u^2=x^2+2x^3+x^4\), \(u^3=x^3+3x^4+O(x^5)\), and \(u^4=x^4+O(x^5)\). 2. Substitution gives \(P_4(x+x^2)=a_0+a_1x+(a_1+a_2)x^2+(2a_2+a_3)x^3+(a_2+3a_3+a_4)x^4+O(x^5)\). 3. Matching coefficients with \(Q_4\) produces \(a_0=1\), \(a_1=2\), \(a_1+a_2=5\), \(2a_2+a_3=10\), and \(a_2+3a_3+a_4=20\). Solving in order gives \(a_2=3\), \(a_3=4\), and \(a_4=5\). 4. Therefore, \(P_4(u)=1+2u+3u^2+4u^3+5u^4\). At \(u=0.1\), \(P_4(0.1)=1+0.2+0.03+0.004+0.0005=1.2345\). 5. Every omitted term of the series for \(f\) contains \(u^k\) with \(k\geq5\). Because \(u=x+x^2\) has lowest degree \(1\), the lowest degree in \(u^k\) is \(x^k\). Such terms begin at degree \(5\) and cannot affect coefficients through degree \(4\).

a) \(a_0=1\), \(a_1=2\), \(a_1+a_2=5\), \(2a_2+a_3=10\), and \(a_2+3a_3+a_4=20\). b) \(P_4(u)=1+2u+3u^2+4u^3+5u^4\). c) \(f(0.1)\approx1.2345\) using the degree-\(4\) Maclaurin polynomial. d) A term \(u^k\) with \(k\geq5\) begins with \(x^k\) after substituting \(u=x+x^2\), so it contributes no term below degree \(5\).