- Start with the general form of a cubic polynomial. - Use the point and inflection conditions at the origin. - A horizontal tangent has slope \(0\). - Solve the resulting system for the remaining coefficients.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Since the origin is on the graph, \(f(0)=0\), so \(d=0\). 2. Since the origin is an inflection point, \(f''(0)=0\). Because \(f''(x)=6ax+2b\), this gives \(b=0\). Thus, \(f(x)=ax^3+cx\). 3. The point \(P=(1, 4)\) gives \(a+c=4\). A horizontal tangent at \(P\) gives \(f'(1)=0\), so \(3a+c=0\). 4. Solving the system gives \(a=-2\) and \(c=6\). Therefore, \(f(x)=-2x^3+6x\). Since \(f'''(x)=-12\ne0\), the origin is indeed an inflection point.

\(f(x)=-2x^3+6x\)

- Use the inflection point at the origin to determine two coefficients. - The tangent-line slope gives the first derivative at the origin. - Use the additional x-intercept to determine the remaining coefficient. - Check that the third derivative is nonzero.

1. Let \(f(x)=ax^3+bx^2+cx+d\). An inflection point at the origin gives \(f(0)=0\) and \(f''(0)=0\), so \(d=0\) and \(b=0\). 2. The tangent line has slope \(-3\), so \(f'(0)=-3\), giving \(c=-3\). 3. The zero at \(x=3\) gives \(27a-9=0\), so \(a=\frac{1}{3}\). 4. Therefore, \(f(x)=\frac{1}{3}x^3-3x\). Since \(f'''(x)=2\ne0\), the origin is indeed an inflection point.

\(f(x)=\frac{1}{3}x^3-3x\)

- How do the given conditions constrain the four unknown coefficients? - Could any one condition rule out several possible coefficients at once? - How can you confirm that your resulting function satisfies every stated condition? - What distinguishes an actual inflection point from a mere zero of the second derivative?

1. Differentiate the proposed cubic: \(f'(x)=3ax^2+2bx+c\) and \(f''(x)=6ax+2b\). 2. Matching \(6ax+2b=6x\) gives \(a=1\) and \(b=0\). 3. The values at zero give \(c=-1\) and \(d=2\), so \(f(x)=x^3-x+2\). 4. Its second derivative changes from negative to positive at \(x=0\), giving the inflection point \((0,2)\).

\(a=1\), \(b=0\), \(c=-1\), \(d=2\). Thus \(f(x)=x^3-x+2\), with inflection point \((0,2)\).

- Think of a familiar cubic that flattens but keeps increasing. - Moving a curve up or down changes where it crosses the y-axis but not its slopes. - Check a horizontal tangent and a concavity change separately.

1. One choice is \(f(x)=x^3+2\), which passes through \((0,2)\). 2. Its derivative is \(f'(x)=3x^2\), nonnegative everywhere and positive except at \(0\), so the function is strictly increasing. 3. Its second derivative is \(f''(x)=6x\), which changes from negative to positive at \(0\), giving an inflection point. 4. Since \(f'(0)=0\), the tangent at that point is horizontal.

One example is \(f(x)=x^3+2\), with \(f'(x)=3x^2\) and \(f''(x)=6x\).

- Start with the general form of a polynomial that has the required symmetry. - Translate the local-extremum conditions into conditions on the first and second derivatives. - You may choose simple coefficient values as long as all conditions are satisfied. - How many zeros must the first derivative have to produce the stated number of extrema?

1. For a), symmetry about the y-axis calls for even powers. Choose \(f(x)=x^4-2x^2\). Then \(f'(x)=4x(x^2-1)\) has critical numbers \(-1,0,1\). Since \(f''(x)=12x^2-4\), \(f''(0)=-4<0\) gives a maximum at \(0\), while \(f''(\pm1)=8>0\) gives minima at \(\pm1\). 2. For b), symmetry about the origin calls for odd powers. Choose \(f(x)=x^3-12x\). Then \(f'(x)=3x^2-12\) vanishes at \(x=-2\), and \(f''(-2)=-12<0\), so it has a local maximum there.

a) \(f(x)=x^4-2x^2\) b) \(f(x)=x^3-12x\)

- A stationary inflection point is an inflection point with a horizontal tangent. - Begin with a translated version of the basic cubic function. - Symmetry about the y-axis requires only even powers of \(x\). - Make the first derivative have exactly the required number of distinct real zeros. - Use the second derivative test to classify the three critical numbers in part b.

1. For part a, a translated cubic power function has a stationary inflection point at its horizontal shift. Choose \(g(x)=(x-2)^3\). Then \(g'(x)=3(x-2)^2\ge0\), with equality only at \(x=2\), so \(g\) is strictly increasing. Also, \(g''(2)=0\), and the concavity changes at \(x=2\). 2. For part b, symmetry about the y-axis requires an even polynomial, so use \(g(x)=ax^4+bx^2+c\). Passing through \((0, 1)\) gives \(c=1\). Choose \(a=1\) and \(b=-4\), giving \(g(x)=x^4-4x^2+1\). Its derivative is \(g'(x)=4x(x^2-2)\), which has three distinct zeros, \(x=0\) and \(x=\pm\sqrt{2}\). Also, \(g''(x)=12x^2-8\). Since \(g''(0)=-8<0\), \(x=0\) is a local maximum; since \(g''(\pm\sqrt{2})=16>0\), \(x=\pm\sqrt{2}\) are local minima. Thus the graph has exactly three local extrema.

a) One possible function is \(g(x)=(x-2)^3\). b) One possible function is \(g(x)=x^4-4x^2+1\).

- What powers can appear in a polynomial whose graph is symmetric about the y-axis? - Use the derivative tests for a local maximum and for inflection points. - Choose simple coefficients that make the second derivative have two distinct real zeros. - Verify every condition after choosing the function.

1. A polynomial symmetric about the y-axis has only even powers, so start with \(f(x)=ax^4+bx^2+c\). 2. Choose \(f(x)=x^4-6x^2\), an even fourth-degree polynomial. 3. The derivatives are \(f'(x)=4x^3-12x\) and \(f''(x)=12x^2-12\). Since \(f'(0)=0\) and \(f''(0)=-12<0\), the second derivative test gives a local maximum at \(0\). 4. The second derivative has exactly two zeros, \(x=\pm1\), and changes sign at both. Hence these are exactly the inflection-point x-coordinates.

One possible function is \(f(x)=x^4-6x^2\).

- A known extremum gives information about both the function value and the derivative. - Use \(f(1)=2\) and \(f'(1)=0\) to form two equations. - Use the sign of the second derivative to classify the extremum.

1. Differentiate: \(f'(x)=2ax+b\). Since \(x=1\) is a critical number, \(f'(1)=0\) gives \(2a+b=0\). 2. The point \((1,2)\) lies on the graph, so \(a+b+3=2\), or \(a+b=-1\). Solving with \(2a+b=0\) gives \(a=1\), \(b=-2\). 3. The second derivative is \(f''(x)=2a=2\). In particular, \(f''(1)=2>0\), so the second derivative test confirms a local minimum at \((1,2)\).

a) \(f'(x)=2ax+b\) b) \(a=1\), \(b=-2\) c) Local minimum at \((1,2)\)

- Use the point condition and the second-derivative condition to form two equations. - Check every zero of the second derivative for a concavity change. - Find all zeros of the first derivative and test for sign changes. - A repeated zero of the first derivative may not produce an extremum.

1. Since \(f(1)=-6\), \(a+b=-6\). Also, \(f''(x)=12ax^2+6bx\). The necessary inflection condition \(f''(2)=0\) gives \(48a+12b=0\). Solving yields \(a=2\) and \(b=-8\), so \(f(x)=2x^4-8x^3\). 2. Then \(f''(x)=24x(x-2)\) changes sign at both \(x=0\) and \(x=2\). The inflection points are \((0,0)\) and \((2,-32)\). 3. The first derivative is \(f'(x)=8x^2(x-3)\), so the critical numbers are \(0\) and \(3\). Since \(f''(0)=0\), the second derivative test is inconclusive at \(0\). The first derivative is negative on both sides of \(0\), so this point is a stationary inflection point, not a local extremum. 4. At \(x=3\), \(f''(3)=72>0\), so the second derivative test gives a local minimum. Since \(f(3)=-54\), the minimum is \((3,-54)\).

a) \(a=2\), \(b=-8\) b) \((0,0)\) and \((2,-32)\) c) Local minimum: \((3,-54)\). The point \((0,0)\) is a stationary inflection point, not an extremum.

- Apply the necessary condition for an inflection point twice. - Solve the resulting system of two equations. - Factor the first derivative and use the discriminant to count real critical numbers. - Use the second derivative to classify the critical point.

1. The second derivative is \(g''(x)=12x^2+6ax+2b\). Setting \(g''(1)=g''(2)=0\) gives \(12+6a+2b=0\) and \(48+12a+2b=0\), so \(a=-6\) and \(b=12\). 2. Then \(g''(x)=12(x-1)(x-2)\), which changes sign at both specified inputs, verifying the inflection points. 3. The function is \(g(x)=x^4-6x^3+12x^2\), and \(g'(x)=2x(2x^2-9x+12)\). The quadratic factor has discriminant \(81-96=-15<0\), so the only real critical number is \(x=0\). 4. Since \(g''(0)=24>0\), the second derivative test gives a local minimum at \(x=0\). Also, \(g(0)=0\), so the only local extremum is \((0,0)\).

a) \(a=-6\), \(b=12\) b) Only local extremum: local minimum at \((0,0)\)

- What equations follow from the origin being an inflection point on the graph? - At the point of tangency, the function and tangent line have the same value and slope. - Use the inflection-point conditions to reduce the number of unknown coefficients. - Solve the resulting linear system.

1. Write a general cubic polynomial: \(f(x)=ax^3+bx^2+cx+d\). 2. Since the origin is on the graph, \(f(0)=0\), so \(d=0\). Also, \(f''(x)=6ax+2b\), and the inflection point at \(x=0\) gives \(f''(0)=0\), so \(b=0\). Thus \(f(x)=ax^3+cx\). 3. The tangent line has value \(-3(2)+10=4\) at \(x=2\), so \(f(2)=4\). Its slope is \(-3\), so \(f'(2)=-3\). 4. These conditions give \(8a+2c=4\) and \(12a+c=-3\). 5. Solving the system gives \(a=-\frac{5}{8}\) and \(c=\frac{9}{2}\). 6. Therefore, \(f(x)=-\frac{5}{8}x^3+\frac{9}{2}x\). Since \(f'''(x)=-\frac{15}{4}\neq0\), the origin is indeed an inflection point.

\(f(x)=-\frac{5}{8}x^3+\frac{9}{2}x\)

- Use the center point to find \(A\). - Substitute the second known point to solve for \(k\). - The most extreme slope occurs at an inflection point. - Multiply the decimal slope magnitude by \(100\%\) to obtain percent grade.

1. Since \(f(0)=A=12\), \(A=12\). Using \(f(20)=8\), \(12e^{-400k}=8\), so \(k=\frac{\ln(1.5)}{400}\approx0.001014\,\text{m}^{-2}\). 2. The steepest slope occurs at an inflection point. Since \(f''(x)=12e^{-kx^2}(-2k+4k^2x^2)\), the positive inflection point satisfies \(x^2=\frac{1}{2k}\). Thus, \(x=\sqrt{\frac{200}{\ln(1.5)}}\approx22.21\). At this point, \(f(x)=12e^{-1/2}\approx7.28\). The point is approximately \((22.21, 7.28)\). 3. The first derivative is \(f'(x)=-2kxf(x)\). At the inflection point, \(f'(x)\approx-0.328\), which corresponds to a downhill grade of about \(32.8\%\).

a) \(A=12\), \(k=\frac{\ln(1.5)}{400}\approx0.001014\,\text{m}^{-2}\) b) Approximately \((22.21, 7.28)\) c) About \(32.8\%\) downhill

- Use y-axis symmetry to determine which powers of \(x\) may appear. - Translate the intercept information into equations. - Use the second derivative for the inflection-point condition. - Solve the resulting system for the coefficients.

1. Symmetry about the y-axis means the polynomial has only even powers, so write \(f(x)=ax^4+bx^2+c\). 2. The y-intercept gives \(f(0)=4\), so \(c=4\). 3. The x-intercept gives \(f(2)=16a+4b+4=0\). 4. Since \(f''(x)=12ax^2+2b\), the inflection-point condition \(f''(1)=0\) gives \(12a+2b=0\), or \(b=-6a\). 5. Substitute into the intercept equation: \(16a+4(-6a)+4=0\), so \(a=\frac12\) and \(b=-3\). 6. Thus, \(f(x)=\frac12x^4-3x^2+4\). Also, \(f'''(1)=12\ne0\), confirming an inflection point at \(x=1\).

\(f(x)=\frac12x^4-3x^2+4\)

- Which polynomial powers are allowed by y-axis symmetry? - Use both the given point and the horizontal tangent to find the coefficients. - Determine the sign of the second derivative at \(x=1\).

1. Symmetry gives \(f_k(x)=ax^4+bx^2+c\). Passing through the origin gives \(c=0\). 2. The point and the horizontal-tangent condition give \(a+b=k\) and \(4a+2b=0\). Thus \(a=-k\) and \(b=2k\), so \(f_k(x)=-kx^4+2kx^2\). 3. Since \(f_k''(x)=-12kx^2+4k\), \(f_k''(1)=-8k\). For \(k>0\), the point is a local maximum; for \(k<0\), a local minimum.

a) \(f_k(x)=-kx^4+2kx^2\) b) Local maximum for \(k>0\); local minimum for \(k<0\).

- Use y-axis symmetry to simplify the general fourth-degree polynomial. - Translate the y-intercept, point, slope, and inflection-point information into equations. - Solve the resulting system for the coefficients. - Verify the inflection point with the third derivative.

1. Symmetry about the y-axis gives \(f(x)=ax^4+bx^2+c\). The y-intercept gives \(c=2.5=\frac52\). 2. The point, slope, and inflection-point conditions at \(x=1\) give \(a+b+\frac52=0\), \(4a+2b=-4\), and \(12a+2b=0\). 3. The third equation gives \(b=-6a\). Substitute into the second equation: \(4a-12a=-4\), so \(a=\frac12\) and \(b=-3\). Therefore, \(f(x)=\frac12x^4-3x^2+\frac52\). The point condition is satisfied, and \(f'''(1)=12\ne0\), confirming the inflection point.

\(f(x)=\frac12x^4-3x^2+\frac52\)

- Use the point and the horizontal-tangent condition to write two equations in \(a\) and \(b\). - Recall that \(\ln(e)=1\). - Compare the growth of a logarithm with the growth of a quadratic term. - Use the second derivative test.

1. Since \(P=(e,1)\) is on the graph, \(a+be^2=1\). 2. The derivative is \(g'(x)=\frac{a}{x}+2bx\). The horizontal tangent at \(x=e\) gives \(\frac{a}{e}+2be=0\), or \(a+2be^2=0\). 3. Subtracting the first equation from the second gives \(be^2=-1\), so \(b=-\frac{1}{e^2}\). Substitution gives \(a=2\). 4. Therefore, \(g(x)=2\ln(x)-\frac{x^2}{e^2}\). As \(x\to0^+\), the logarithmic term approaches \(-\infty\), so \(g(x)\to-\infty\). As \(x\to\infty\), the negative quadratic term dominates, so \(g(x)\to-\infty\). 5. The second derivative is \(g''(x)=-\frac{2}{x^2}-\frac{2}{e^2}\). Thus \(g'(e)=0\) and \(g''(e)=-\frac{4}{e^2}<0\), so \(P\) is a local maximum by the second derivative test.

a) \(a=2\), \(b=-\frac{1}{e^2}\). b) \(\lim_{x\to0^+}g(x)=-\infty\) and \(\lim_{x\to\infty}g(x)=-\infty\). c) Since \(g'(e)=0\) and \(g''(e)=-\frac{4}{e^2}<0\), \(P\) is a local maximum.

- Start with the general form of a cubic polynomial. - Translate the inflection-point coordinates into conditions on \(f\) and \(f''\). - A horizontal tangent has slope \(0\). - Use the coordinates of each point to form equations.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Then \(f'(x)=3ax^2+2bx+c\) and \(f''(x)=6ax+2b\). 2. Since \(W(0, 2)\) is an inflection point, \(f(0)=2\) and \(f''(0)=0\). Thus, \(d=2\) and \(b=0\). 3. Since \(P(2, 6)\) lies on the graph, \(f(2)=6\), which gives \(4a+c=2\). The tangent at \(x=2\) is horizontal, so \(f'(2)=0\), which gives \(12a+c=0\). 4. Solving the system gives \(a=-\frac{1}{4}\) and \(c=3\). Therefore, \(f(x)=-\frac{1}{4}x^3+3x+2\). Its third derivative is \(-\frac{3}{2}\ne0\), confirming the inflection point.

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

- Use only even powers in the general quartic form. - Use the y-intercept and the given point to form two equations. - At a local extremum, the first derivative is \(0\). - Verify the local minimum with the second derivative.

1. Because \(f\) is an even quartic polynomial, let \(f(x)=ax^4+cx^2+e\). Then \(f'(x)=4ax^3+2cx\). 2. The y-intercept gives \(f(0)=5\), so \(e=5\). The point \(P=(1,-2)\) gives \(a+c=-7\). 3. Since \(x=2\) is a local minimum, \(f'(2)=0\). Thus \(32a+4c=0\), or \(8a+c=0\). 4. Solving gives \(a=1\) and \(c=-8\). Therefore, \(f(x)=x^4-8x^2+5\). 5. Its second derivative is \(f''(x)=12x^2-16\), so \(f''(2)=32>0\), confirming that \(x=2\) is a local minimum.

\(f(x)=x^4-8x^2+5\).

- Begin with the general form of a cubic polynomial. - Translate the local-minimum coordinates into conditions on \(f\) and \(f'\). - Translate the inflection-point coordinates into conditions on \(f\) and \(f''\). - Verify both point types after finding the coefficients.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Then \(f'(x)=3ax^2+2bx+c\) and \(f''(x)=6ax+2b\). 2. The local minimum at \(T=(0,2)\) gives \(f(0)=2\) and \(f'(0)=0\). Therefore, \(d=2\) and \(c=0\). 3. The inflection point at \(W=(1,4)\) gives \(f(1)=4\) and \(f''(1)=0\). Thus \(a+b=2\) and \(6a+2b=0\). 4. Solving gives \(a=-1\), \(b=3\), so \(f(x)=-x^3+3x^2+2\). 5. Since \(f''(0)=6>0\), the second derivative test confirms the minimum at \(T\). Also \(f'''(x)=-6\ne0\), confirming the concavity change at \(W\).

\(f(x)=-x^3+3x^2+2\).

- An even quartic contains only even powers of \(x\). - A point on the graph gives an equation involving \(f\). - A local extremum gives both a function-value condition and a first-derivative condition. - Use the second derivative to determine its type.

1. Because the polynomial is even, let \(f(x)=ax^4+cx^2+e\). 2. Since \(P=(0,4)\) is on the graph, \(e=4\). The point \(E=(2,0)\) gives \(16a+4c+4=0\), or \(4a+c=-1\). 3. The extremum condition \(f'(2)=0\), with \(f'(x)=4ax^3+2cx\), gives \(32a+4c=0\), or \(8a+c=0\). 4. Solving gives \(a=\frac{1}{4}\), \(c=-2\), so \(f(x)=\frac{1}{4}x^4-2x^2+4\). 5. Since \(f''(x)=3x^2-4\) and \(f''(2)=8>0\), \(E=(2,0)\) is a local minimum.

\(f(x)=\frac{1}{4}x^4-2x^2+4\). The extremum at \((2,0)\) is a local minimum.

- Start with the general cubic form and its first two derivatives. - Translate the y-intercept and slope information into equations. - An inflection point gives a function-value condition and a second-derivative condition. - Solve the resulting system of equations.

1. Let \(f(x)=ax^3+bx^2+cx+d\). The y-intercept gives \(f(0)=3\), so \(d=3\). The slope at the y-intercept gives \(f'(0)=-6\), so \(c=-6\). 2. Since \(W=(1, 1)\) is an inflection point, \(f(1)=1\) and \(f''(1)=0\). These conditions give \(a+b=4\) and \(3a+b=0\). 3. Solving the system gives \(a=-2\) and \(b=6\). Therefore, \(f(x)=-2x^3+6x^2-6x+3\). The nonzero third derivative confirms the inflection point.

\(f(x)=-2x^3+6x^2-6x+3\)

- Begin with the general form of a cubic polynomial. - An inflection point gives conditions on the function and second derivative. - A local minimum gives conditions on the function and first derivative. - Verify the point classifications after finding the coefficients.

1. Let \(f(x)=ax^3+bx^2+cx+d\). The inflection point at \(W=(0,1)\) gives \(f(0)=1\) and \(f''(0)=0\), so \(d=1\) and \(b=0\). 2. The minimum at \(T=(1,0)\) gives \(f(1)=0\) and \(f'(1)=0\). Therefore \(a+c=-1\) and \(3a+c=0\). 3. Solving gives \(a=\frac{1}{2}\) and \(c=-\frac{3}{2}\), so \(f(x)=\frac{1}{2}x^3-\frac{3}{2}x+1\). 4. Since \(f''(1)=3>0\), \(T\) is a local minimum by the second derivative test. Also \(f''(0)=0\) and \(f'''(x)=3\ne0\), so the concavity changes at \(W\).

\(f(x)=\frac{1}{2}x^3-\frac{3}{2}x+1\).

- The tangent-line equation gives both the function value and slope at \(x=1\). - Use the second-derivative condition for the inflection point. - Use the origin to determine the constant term. - Solve the resulting linear system.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Since the graph passes through the origin, \(d=0\). 2. At \(x=1\), the tangent line has value \(1\) and slope \(-1\). Therefore, \(f(1)=1\) and \(f'(1)=-1\). The inflection condition gives \(f''(1)=0\). 3. These conditions produce \(a+b+c=1\), \(3a+2b+c=-1\), and \(6a+2b=0\). 4. Solving the system gives \(a=2\), \(b=-6\), and \(c=5\). Therefore, \(f(x)=2x^3-6x^2+5x\). Since \(f'''(x)=12\ne0\), \(x=1\) is indeed an inflection-point x-coordinate.

\(f(x)=2x^3-6x^2+5x\)

- Use only even powers for an even quartic polynomial. - Use the y-intercept and the given point to form equations. - At a local minimum, the first derivative is \(0\). - Verify the classification with the second derivative.

1. Because the polynomial is even, let \(f(x)=ax^4+bx^2+c\). The y-intercept gives \(c=2\). 2. The point \(P(2, -6)\) gives \(16a+4b+2=-6\), or \(16a+4b=-8\). 3. Since \(x=2\) is a local minimum, \(f'(2)=0\). Because \(f'(x)=4ax^3+2bx\), this gives \(32a+4b=0\). 4. Solving the system gives \(a=\frac{1}{2}\) and \(b=-4\). Therefore, \(f(x)=\frac{1}{2}x^4-4x^2+2\). 5. Since \(f''(x)=6x^2-8\), \(f''(2)=16>0\), confirming the local minimum.

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

- Use only even powers for an even quartic polynomial. - The second derivative is \(0\) at an inflection point. - The first derivative gives the tangent slope. - Solve the resulting system for the coefficients.

1. Because the polynomial is even, let \(f(x)=ax^4+bx^2+c\). The point \(S=(0, 5)\) gives \(c=5\). 2. The inflection condition \(f''(1)=0\) gives \(12a+2b=0\). The tangent slope condition \(f'(1)=-4\) gives \(4a+2b=-4\). 3. Solving the system gives \(a=\frac{1}{2}\) and \(b=-3\). Therefore, \(f(x)=\frac{1}{2}x^4-3x^2+5\). 4. Since \(f'''(1)=12\ne0\), \(x=1\) is indeed an inflection point.

\(f(x)=\frac{1}{2}x^4-3x^2+5\)

- Translate a stationary inflection point into conditions on \(f\), \(f'\), and \(f''\). - Conditions at the origin simplify the general quartic form. - A horizontal tangent has slope \(0\). - Verify the concavity change with the third derivative.

1. Let \(f(x)=ax^4+bx^3+cx^2+dx+e\). A stationary inflection point at the origin gives \(f(0)=f'(0)=f''(0)=0\), so \(e=d=c=0\). Thus, \(f(x)=ax^4+bx^3\). 2. The point \(P=(1, 2)\) gives \(a+b=2\). The horizontal tangent at \(x=\frac{3}{2}\) gives \(f'(\frac{3}{2})=0\), so \(\frac{27}{2}a+\frac{27}{4}b=0\). 3. Solving the system gives \(a=-2\) and \(b=4\). Therefore, \(f(x)=-2x^4+4x^3\). 4. Since \(f'''(0)=24\ne0\), the origin is a stationary inflection point.

\(f(x)=-2x^4+4x^3\)

- Begin with the general cubic form. - Translate the point, extremum, and inflection conditions into equations. - Use the second-derivative condition first to simplify the system. - Classify the extremum after finding the polynomial.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Passing through the origin gives \(d=0\). 2. The extremum at \(E=(2,-4)\) gives \(f(2)=-4\) and \(f'(2)=0\); the inflection condition gives \(f''(1)=0\). 3. From \(f''(1)=6a+2b=0\), \(b=-3a\). Substituting into \(f'(2)=12a+4b+c=0\) gives \(c=0\). 4. Using \(f(2)=-4\) gives \(8a+4(-3a)=-4\), so \(a=1\), \(b=-3\). Therefore \(f(x)=x^3-3x^2\). 5. Since \(f''(2)=6>0\), \(E=(2,-4)\) is a local minimum. Also \(f'''(x)=6\ne0\) confirms the inflection at \(x=1\).

\(f(x)=x^3-3x^2\). The extremum at \(E=(2,-4)\) is a local minimum.

- What must the first and second derivatives equal at a stationary inflection point? - Use those conditions to form a linear system in \(a\) and \(b\). - Check a higher derivative to confirm the change in concavity.

1. A stationary inflection point at \(x=1\) requires \(f'(1)=0\) and \(f''(1)=0\). 2. The derivatives are \(f'(x)=4x^3+3ax^2+2bx\) and \(f''(x)=12x^2+6ax+2b\). 3. Substituting \(x=1\) gives \(3a+2b=-4\) and \(6a+2b=-12\). 4. Subtracting the equations gives \(3a=-8\), so \(a=-\frac{8}{3}\). Substitution gives \(b=2\). 5. The third derivative is \(f'''(x)=24x+6a\), and \(f'''(1)=8\ne0\). Therefore, the concavity changes at \(x=1\), confirming a stationary inflection point.

\(a=-\frac{8}{3}\), \(b=2\)

- How is the annual increase \(G(x)\) related to the reproduction model \(f(x)\)? - What derivative condition holds at an interior maximum? - How can the population value for the following year be written as an equation involving \(f\)? - Use the two conditions to create a system for \(a\) and \(b\).

1. Write the annual-increase function: \(G(x)=f(x)-x=ax^2+(b-1)x\). 2. Since \(f(400)=500\), \(160{,}000a+400b=500\). 3. The increase is maximized at \(x=400\), so \(G'(400)=0\). Because \(G'(x)=2ax+b-1\), this gives \(800a+b=1\). 4. From the second equation, \(b=1-800a\). Substitute into the first equation: \(160{,}000a+400(1-800a)=500\). 5. Simplify: \(-160{,}000a=100\), so \(a=-\frac{1}{1600}=-0.000625\). 6. Then \(b=1-800\left(-\frac{1}{1600}\right)=\frac{3}{2}=1.5\). 7. Since \(G''(x)=2a=-\frac{1}{800}<0\), the critical point is a maximum, as required.

\(a=-\frac{1}{1600}=-0.000625\) and \(b=\frac{3}{2}=1.5\)

- Use the required function value and the condition that the first derivative is zero. - Solve the resulting system for the two parameters. - Use the second derivative test to classify the critical point.

1. The point \((-2, -1)\) must satisfy \(f_{a,b}(-2)=-1\), and a local extremum requires \(f_{a,b}^{\prime}(-2)=0\). 2. From the function value, \(\frac{-2a+b}{5}=-1\), so \(-2a+b=-5\) and \(b=2a-5\). 3. Differentiate: \(f_{a,b}^{\prime}(x)=\frac{a(x^2+1)-2x(ax+b)}{(x^2+1)^2}=\frac{-ax^2-2bx+a}{(x^2+1)^2}\). 4. Applying \(f_{a,b}^{\prime}(-2)=0\) gives \(-3a+4b=0\). Substitute \(b=2a-5\): \(-3a+4(2a-5)=0\), so \(5a=20\) and \(a=4\). Then \(b=3\). 5. For \(a=4\) and \(b=3\), the numerator of \(f^{\prime}\) is \(N(x)=-4x^2-6x+4\). Since \(N(-2)=0\), \(f^{\prime\prime}(-2)=\frac{N^{\prime}(-2)}{25}=\frac{10}{25}=0.4>0\). Therefore, \((-2, -1)\) is a local minimum.

\(a=4\) and \(b=3\). Since \(f^{\prime\prime}(-2)=0.4>0\), the point \((-2, -1)\) is a local minimum.

- Read the three marked points carefully. - At each marked local extremum, the first derivative is \(0\). - Use function-value and derivative conditions at extrema and inflection points. - The inflection point on the y-axis simplifies the cubic form. - A local extremum on the x-axis is a multiple zero.

1. From the graph, \(H=(-1, 4)\), \(T=(1, 0)\), and \(W=(0, 2)\). 2. Let \(f(x)=ax^3+bx^2+cx+d\). Since \(W=(0, 2)\) is an inflection point, \(f(0)=2\) and \(f''(0)=0\), so \(d=2\) and \(b=0\). 3. Using \(T=(1, 0)\), the equations \(f(1)=0\) and \(f'(1)=0\) give \(a+c=-2\) and \(3a+c=0\). Thus, \(a=1\) and \(c=-3\), so \(f(x)=x^3-3x+2\). 4. Factoring gives \(f(x)=(x-1)^2(x+2)\). Therefore, \(x=1\) is a double zero and \(x=-2\) is a simple zero.

a) \(H=(-1, 4)\), \(T=(1, 0)\), \(W=(0, 2)\) b) \(f(x)=x^3-3x+2\) c) \(x=1\), multiplicity \(2\); \(x=-2\), multiplicity \(1\)

- Read the exact coordinates of the labeled extrema from the graph. - At each extremum, use both the point and zero-slope conditions. - Substitute \(x=2\) to test Hana's point. - Find the midpoint of the two extrema.

1. From the graph, \(H=(-1, 4)\) and \(T=(1, 0)\). The extrema give \(f(-1)=4\), \(f(1)=0\), \(f'(-1)=0\), and \(f'(1)=0\). Solving these four equations gives \(a=1\), \(b=0\), \(c=-3\), and \(d=2\). Thus, \(f(x)=x^3-3x+2\). 2. Evaluating at \(x=2\) gives \(f(2)=8-6+2=4\ne5\), so \(P=(2, 5)\) is not on the graph. 3. The graph shows \(f(2)=4\) and \(W=(0, 2)\). The midpoint of \(H=(-1, 4)\) and \(T=(1, 0)\) is \((0, 2)\), so the extrema are paired by a half-turn about the inflection point.

a) \(f(x)=x^3-3x+2\), and \(f(2)=4\ne5\), so the claim is false. b) \(f(2)=4\) and \(W=(0, 2)\). The point \(W\) is the midpoint of \(H\) and \(T\), matching the graph's point symmetry.

- Read the coordinates of the marked local maximum and local minimum. - Count the turning points shown in the graph. - Relate the maximum possible number of turning points to polynomial degree. - Use origin symmetry to determine which powers may appear. - At the marked local minimum, use both the point and zero-slope conditions.

1. The graph has two local extrema. A degree-\(n\) polynomial can have at most \(n-1\) local extrema, so the least possible degree is \(3\). 2. The graph is symmetric about the origin, so a cubic of the least possible degree has the form \(f(x)=ax^3+cx\). 3. Using the local minimum \(T=(2,-4)\), the conditions \(f(2)=-4\) and \(f'(2)=0\) give \(4a+c=-2\) and \(12a+c=0\). Solving gives \(a=\frac14\) and \(c=-3\). Therefore, \(f(x)=\frac14x^3-3x\).

a) Degree \(3\) b) \(f(x)=ax^3+cx\) c) \(f(x)=\frac{1}{4}x^3-3x\)

- Smooth connections to horizontal platforms require matching heights and zero slopes. - Begin with a general cubic polynomial and its derivative. - Use the four endpoint conditions to determine the cubic. - The extremum of the slope occurs where the second derivative is \(0\). - Compare the maximum derivative value with \(0.6\).

1. Smooth horizontal connections give \(f(0)=0\), \(f'(0)=0\), \(f(4)=2\), and \(f'(4)=0\). Solving for a cubic gives \(f(x)=-\frac{1}{16}x^3+\frac{3}{8}x^2\). 2. The second derivative is \(f''(x)=-\frac{3}{8}x+\frac{3}{4}\), so the inflection point occurs at \(x=2\). Since \(f(2)=1\), the point is \(W=(2, 1)\). This is where the ramp is steepest. 3. The maximum slope is \(f'(2)=\frac{3}{4}=0.75\), or \(75\%\). Since \(75\%>60\%\), the design does not meet the requirement.

a) \(f(x)=-\frac{1}{16}x^3+\frac{3}{8}x^2\) b) \(W=(2, 1)\); the ramp is steepest there. c) No. The maximum grade is \(75\%\), which exceeds \(60\%\).

- Use the inflection point on the y-axis to find the constant term. - At a marked local extremum, use both its coordinates and the condition that the derivative is \(0\).

1. The graph's inflection point is \((0, 1)\), so \(c=1\). 2. The graph has a local maximum at \((-1, 3)\) and a local minimum at \((1, -1)\). Since \(f'(x)=3ax^2+b\), the zero slope at \(x=1\) gives \(3a+b=0\), so \(b=-3a\). 3. Using \(f(1)=-1\), \(a-3a+1=-1\), so \(a=1\) and \(b=-3\). 4. Therefore, \(f(x)=x^3-3x+1\).

\(f(x)=x^3-3x+1\)

- Use the zeros to write a factored cubic with an unknown leading factor. - Substitute the additional point to find the factor. - Use the second and third derivatives for the inflection point. - Test \(f(-x)\) for origin symmetry. - A shift right by \(2\) replaces \(x\) with \(x-2\), and a shift up by \(1\) adds \(1\).

1. The zeros give \(f(x)=a x(x+3)(x-3)=a(x^3-9x)\). Using \(P=(1, -2)\), \(-2=-8a\), so \(a=\frac{1}{4}\). Thus, \(f(x)=\frac{1}{4}(x^3-9x)\). 2. The derivatives are \(f'(x)=\frac{3}{4}x^2-\frac{9}{4}\), \(f''(x)=\frac{3}{2}x\), and \(f'''(x)=\frac{3}{2}\). Since \(f''(0)=0\) and \(f'''(0)\ne0\), \(W=(0, 0)\) is an inflection point. Its tangent slope is \(f'(0)=-\frac{9}{4}\), so the tangent line is \(y=-\frac{9}{4}x\). 3. Since \(f(-x)=-f(x)\), the graph is symmetric about the origin. Translating right \(2\) and up \(1\) gives \(h(x)=f(x-2)+1=\frac{1}{4}((x-2)^3-9(x-2))+1\), whose inflection point is \(W'=(2,1)\).

a) \(f(x)=\frac{1}{4}(x^3-9x)\) b) \(W=(0, 0)\); tangent line: \(y=-\frac{9}{4}x\) c) \(f(-x)=-f(x)\); \(h(x)=\frac{1}{4}((x-2)^3-9(x-2))+1\)

- Translate the zero slope and local maximum into derivative conditions. - Start with the general form of a cubic polynomial. - Use the starting point to simplify the formula. - How can you check that the hilltop is a maximum after finding the coefficients?

1. Start with \(f(x)=ax^3+bx^2+cx+d\). 2. From \(f(0)=0\) and \(f'(0)=0\), we get \(d=c=0\). Thus \(f(x)=ax^3+bx^2\). 3. The maximum at \((2,6)\) gives \(f(2)=6\) and \(f'(2)=0\), so \(8a+4b=6\) and \(12a+4b=0\). 4. Solving yields \(a=-\frac{3}{2}\) and \(b=\frac{9}{2}\). Therefore, \(f(x)=-\frac{3}{2}x^3+\frac{9}{2}x^2\). 5. Since \(f''(x)=-9x+9\), \(f''(2)=-9<0\), confirming a local maximum.

\(f(x)=-\frac{3}{2}x^3+\frac{9}{2}x^2\)

- A horizontal tangent gives a condition on the first derivative. - Use the inflection point to find conditions on the function and second derivative. - How do y-values change when reflected across \(y=1\)? - Use the second derivative to classify each extremum.

1. Write \(f(x)=ax^3+bx^2+cx+d\). 2. At the inflection point \(W=(1,1)\), \(f(1)=1\) and \(f''(1)=0\). Since \(f''(x)=6ax+2b\), \(b=-3a\). 3. Since \(f'(0)=0\), the coefficient \(c=0\). 4. Using \(A=(3,5)\), we obtain \(5=27a+9b+d=d\). Also \(f(1)=1\), so \(a-3a+5=1\). Thus \(a=2\), \(b=-6\), and \(f(x)=2x^3-6x^2+5\). 5. Reflection across \(y=1\) maps each \(y\) to \(2-y\). Thus \(g(x)=2-f(x)=-2x^3+6x^2-3\). 6. The critical numbers follow from \(f'(x)=6x(x-2)\): \(x=0\) and \(x=2\). Since \(f''(x)=12x-12\), \(f''(0)=-12<0\) and \(f''(2)=12>0\). Thus \((0,5)\) is a local maximum and \((2,-3)\) a local minimum.

a) \(f(x)=2x^3-6x^2+5\) b) \(g(x)=2-f(x)=-2x^3+6x^2-3\) c) Local maximum: \((0,5)\); local minimum: \((2,-3)\)

- Symmetry about the y-axis restricts the powers that may appear. - An inflection point on the x-axis gives a function-value condition and a second-derivative condition. - Use the y-intercept to determine the constant term. - Solve the resulting two-equation system for the remaining coefficients.

1. Because \(f\) is even, write \(f(x)=ax^4+bx^2+c\). 2. The point \(M=(0, 4)\) gives \(c=4\). At the inflection point with \(x=1\), both \(f(1)=0\) and \(f''(1)=0\). The first condition gives \(a+b=-4\). Since \(f''(x)=12ax^2+2b\), the second gives \(12a+2b=0\), or \(6a+b=0\). Solving the system yields \(a=\frac{4}{5}\) and \(b=-\frac{24}{5}\). Therefore, \(f(x)=\frac{4}{5}x^4-\frac{24}{5}x^2+4\). 3. Since \(f''(0)=-\frac{48}{5}<0\), \(M=(0, 4)\) is a local maximum. Also, \(f'''(\pm1)=\pm\frac{96}{5}\ne0\), so \(x=-1\) and \(x=1\) are indeed inflection-point x-coordinates.

1. \(f(x)=ax^4+bx^2+c\) 2. \(f(x)=\frac{4}{5}x^4-\frac{24}{5}x^2+4\)

- Use the y-axis symmetry to reduce the number of coefficients. - A marked inflection point gives conditions on the function and its second derivative. - Find the critical numbers from the first derivative, then classify the extrema. - Keep \(\sqrt{3}\) as an exact value.

1. Because \(g\) is even, write \(g(x)=ax^4+bx^2+c\). The graph passes through the origin, so \(c=0\). The marked inflection point \(I_2=\left(1, \frac{5}{2}\right)\) gives \(g(1)=\frac{5}{2}\) and \(g''(1)=0\). Therefore, \(a+b=\frac{5}{2}\) and \(12a+2b=0\). Solving gives \(a=-\frac{1}{2}\) and \(b=3\), so \(g(x)=-\frac{1}{2}x^4+3x^2\). Since \(g'''(\pm1)=\mp12\ne0\), both marked points are indeed inflection points. 2. The derivative is \(g'(x)=-2x^3+6x=-2x(x^2-3)\). Thus, the critical numbers are \(x=0\) and \(x=\pm\sqrt{3}\). Since \(g''(0)=6>0\), \(x=0\) is a local minimum. At \(x=\pm\sqrt{3}\), \(g''(x)=-6x^2+6=-12<0\), so both are local maxima. Their common function value is \(g(\pm\sqrt{3})=-\frac{1}{2}(9)+3(3)=\frac{9}{2}\).

1. \(g(x)=-\frac{1}{2}x^4+3x^2\) 2. Local maxima: \((-\sqrt{3}, \frac{9}{2})\) and \((\sqrt{3}, \frac{9}{2})\)

- Use the y-axis symmetry to simplify the polynomial form. - Translate “meets the ground with a horizontal tangent” into a function-value condition and a derivative condition. - Use the center height to determine the constant term. - Solve the resulting system for the remaining coefficients.

1. Because the arch is symmetric about the y-axis, write \(f(x)=ax^4+bx^2+c\). 2. The center point \((0, 20)\) gives \(c=20\). 3. The endpoint \((20, 0)\) gives \(160{,}000a+400b+20=0\). 4. A horizontal tangent at \(x=20\) requires \(f'(20)=0\). Since \(f'(x)=4ax^3+2bx\), \(32{,}000a+40b=0\), so \(b=-800a\). 5. Substitution into the endpoint equation gives \(-160{,}000a=-20\), so \(a=\frac{1}{8000}\) and \(b=-\frac{1}{10}\). 6. Therefore, \(f(x)=\frac{1}{8000}x^4-\frac{1}{10}x^2+20\), for \(-20\le x\le20\). Also, \(f''(0)=-\frac{1}{5}<0\), confirming the maximum at the center.

\(f(x)=\frac{1}{8000}x^4-\frac{1}{10}x^2+20\), for \(-20\le x\le20\)

- Can a curve have a horizontal tangent without turning from rising to falling or vice versa? - Look for an odd-power polynomial that stays increasing through a horizontal tangent. - Change the input and output so the flat point occurs at the requested coordinates. - Check the derivative signs on both sides of that point.

1. A horizontal tangent without a local extremum can occur at a stationary inflection point. 2. Shift the basic cubic \(y=x^3\) left \(3\) units and up \(2\) units: \(f(x)=(x+3)^3+2\). 3. Then \(f(-3)=2\) and \(f'(x)=3(x+3)^2\), so \(f'(-3)=0\). 4. Since \(f'(x)\geq0\) on both sides of \(x=-3\), the derivative does not change sign and no local extremum occurs.

One possible function is \(f(x)=(x+3)^3+2\).

- Convert each width to a radius and each height to units of \(100\,\text{m}\). - Take natural logarithms to solve for the parameters. - The sight line is tangent to the profile at an inflection point. - The top of the tower is the y-intercept of that tangent line.

1. A width of \(200\,\text{m}\) gives radius \(x=1\), and height \(400\,\text{m}\) gives \(h=4\). Similarly, the second condition gives \(x=2\) and \(h=2\). Thus, \(e^{a(b-1)}=4\) and \(e^{a(b-4)}=2\). 2. Taking natural logarithms and subtracting gives \(3a=\ln2\), so \(a=\frac{\ln2}{3}\). Substitution gives \(b=7\). 3. The summit height is \(h(0)=e^{ab}=2^{7/3}\approx5.040\), or about \(504.0\,\text{m}\). 4. The positive inflection point is \(x_W=\sqrt{\frac{1}{2a}}=\sqrt{\frac{3}{2\ln2}}\). Its height is \(h(x_W)=h(0)e^{-1/2}\approx3.057\). 5. Since \(h'(x)=-2axh(x)\), the tangent at \(x_W\) has y-intercept \(-x_Wh'(x_W)+h(x_W)=2h(x_W)\approx6.113\). The tower height is \(6.113-5.040\approx1.073\) units, or about \(107.4\,\text{m}\).

a) \(a=\frac{\ln2}{3}\approx0.231\), \(b=7\) b) About \(107.4\,\text{m}\)

- Translate the point at the origin into a condition on the constant term. - A stationary inflection point provides conditions on the function, first derivative, and second derivative. - A local extremum provides a condition on the first derivative. - Solve the resulting linear system for the coefficients.

1. Write \(f(x)=ax^4+bx^3+cx^2+dx+e\). Passing through the origin gives \(e=0\). 2. A stationary inflection point at \((2, 2)\) gives \(f(2)=2\), \(f'(2)=0\), and \(f''(2)=0\). The local-extremum condition gives \(f'(1)=0\). 3. These conditions produce the system \(16a+8b+4c+2d=2\), \(32a+12b+4c+d=0\), \(48a+12b+2c=0\), \(4a+3b+2c+d=0\). Solving gives \(a=-\frac38\), \(b=\frac52\), \(c=-6\), and \(d=6\). 4. Therefore, \(f(x)=-\frac38x^4+\frac52x^3-6x^2+6x\). Here \(f'''(2)=-3\ne0\), so \((2, 2)\) is an inflection point, and \(f''(1)=-\frac32<0\), so \(x=1\) is a local maximum.

\(f(x)=-\frac38x^4+\frac52x^3-6x^2+6x\)

- Begin with a general cubic polynomial and its first two derivatives. - Use the inflection point at the origin to determine two coefficients. - Use the tangent line to determine the slope at the origin. - Apply the first-derivative condition at \(x=1\). - Use the second derivative to classify the resulting critical point.

1. Let \(f(x)=ax^3+bx^2+cx+d\). An inflection point at the origin gives \(f(0)=0\) and \(f''(0)=0\), so \(d=0\) and \(b=0\). 2. The tangent line \(y=x\) has slope \(1\), so \(f'(0)=1\). Therefore \(c=1\), and \(f(x)=ax^3+x\). 3. A local extremum at \(x=1\) requires \(f'(1)=0\). Since \(f'(x)=3ax^2+1\), this gives \(3a+1=0\), so \(a=-\frac{1}{3}\). 4. Now \(f''(x)=6ax=-2x\), so \(f''(1)=-2<0\). Thus \(x=1\) is a local maximum, not a local minimum. No cubic satisfies all the conditions.

No such cubic polynomial exists. The conditions force \(f(x)=-\frac{1}{3}x^3+x\), which has a local maximum rather than a local minimum at \(x=1\).

- Translate each point and derivative condition into an equation. - Find the only cubic satisfying the point, stationary-point, and inflection conditions. - Use the second derivative to classify the critical point at \(x=2\). - A contradiction in the classification proves nonexistence.

1. Let \(f(x)=ax^3+bx^2+cx+d\). The conditions \(f(0)=0\) and \(f''(3)=0\) give \(d=0\) and \(b=-9a\). 2. The conditions \(f(2)=4\) and \(f'(2)=0\) give \(-28a+2c=4\) and \(-24a+c=0\). 3. Solving yields \(a=\frac{1}{5}\), \(b=-\frac{9}{5}\), and \(c=\frac{24}{5}\). This is the only cubic satisfying the point, critical-point, and inflection conditions. 4. Its second derivative at the critical number is \(f''(2)=-\frac{6}{5}<0\), which means \(T\) is a local maximum, not a local minimum. No such cubic exists.

No such cubic polynomial exists. The conditions force \(f(x)=\frac{1}{5}x^3-\frac{9}{5}x^2+\frac{24}{5}x\), and \(f''(2)=-\frac{6}{5}<0\), so \(T\) is a local maximum.

- Treat the leading coefficient as a free parameter. - Express the remaining coefficients in terms of that parameter. - Evaluate the second derivative at the stationary point. - If it equals zero, what can the third derivative tell you? - Verify that concavity changes at the prescribed inflection x-coordinate.

1. The conditions \(g(0)=0\), \(g''(3)=0\), \(g(2)=4\), and \(g'(2)=0\) give \(e=0\) and equations for the other coefficients. 2. Solving in terms of \(a\) gives \(b=\frac{1}{5}-\frac{42}{5}a\), \(c=\frac{108}{5}a-\frac{9}{5}\), and \(d=\frac{24}{5}-\frac{88}{5}a\). 3. The second derivative at \(x=2\) is \(g''(2)=-\frac{48}{5}a-\frac{6}{5}\). For \(a<-\frac{1}{8}\), it is positive, so \(T\) is a local minimum. For \(a>-\frac{1}{8}\), it is negative and \(T\) is a local maximum. 4. At \(a=-\frac{1}{8}\), \(g''(2)=0\) and the second derivative test is inconclusive. But \(g'''(2)=\frac{3}{2}\ne0\), so the concavity changes at the stationary point and \(T\) is not a local minimum. 5. For \(a<-\frac{1}{8}\), \(g'''(3)=\frac{108}{5}a+\frac{6}{5}\ne0\), so \(x=3\) really is an inflection point. Thus infinitely many quartics satisfy all requirements, precisely for \(a<-\frac{1}{8}\).

Yes. Infinitely many such quartic polynomials exist, with \(a<-\frac{1}{8}\). At \(a=-\frac{1}{8}\), the second derivative test is inconclusive but \(T\) is not a minimum.

- How do the function value and the first two derivatives behave at the specified point? - Consider whether the conditions at \(x=2\) allow you to factor the polynomial before expanding it. - Use the point and slope at \(x=0\) to determine the remaining unknowns. - Check that the second derivative changes sign at the proposed inflection point.

1. A stationary inflection point on the x-axis at \(x=2\) is a zero of multiplicity \(3\). Write \(f(x)=a(x-2)^3(x-k)\). 2. The condition \(f(0)=8\) gives \(8ak=8\), so \(ak=1\). 3. Differentiating gives \(f'(x)=a(x-2)^2(4x-3k-2)\). Thus, \(f'(0)=-12ak-8a=-20\). Using \(ak=1\) gives \(a=1\), and then \(k=1\). 4. Therefore, \(f(x)=(x-2)^3(x-1)=x^4-7x^3+18x^2-20x+8\). Also, \(f'''(2)=6\ne0\), confirming the stationary inflection point.

\(f(x)=x^4-7x^3+18x^2-20x+8\)

- What do the tangent line’s value and slope tell you about \(f(1)\) and \(f'(1)\)? - Translate each graph condition into an equation involving \(f\), \(f'\), or \(f''\). - How many coefficients does a general cubic polynomial have? - Verify the tangent-point condition after finding the coefficients.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Then \(f'(x)=3ax^2+2bx+c\) and \(f''(x)=6ax+2b\). 2. Translate the conditions into equations: \(f(0)=2\), so \(d=2\). \(f(2)=0\), so \(8a+4b+2c=-2\). \(f''(1)=0\), so \(6a+2b=0\). The tangent slope is \(-3\), so \(f'(1)=-3\), giving \(3a+2b+c=-3\). Also, \(f(1)=t(1)=1\), so \(a+b+c=-1\). 3. From \(6a+2b=0\), \(b=-3a\). Substituting into the remaining equations gives \(-2a+c=-1\) and \(-3a+c=-3\). Subtracting yields \(a=2\). Then \(b=-6\), \(c=3\), and \(d=2\). 4. Therefore, \(f(x)=2x^3-6x^2+3x+2\). Verification gives \(f(2)=0\), \(f(1)=1\), \(f'(1)=-3\), and \(f''(1)=0\). Since \(f'''(x)=12\ne0\), the graph changes concavity at \(x=1\).

\(f(x)=2x^3-6x^2+3x+2\)

- Translate each endpoint and horizontal-tangent statement into an equation. - A horizontal tangent means the first derivative is \(0\). - Use the second derivative to find the inflection point. - Convert slope magnitude to percent and use inverse tangent for the angle.

1. The equation \(f(10)=0\) means the slide reaches ground level after \(10\,\text{m}\) horizontally. The equation \(f'(0)=0\) means the slide begins horizontally. 2. Let \(f(x)=ax^3+bx^2+cx+d\). The conditions \(f(0)=8\), \(f'(0)=0\), \(f(10)=0\), and \(f'(10)=0\) give \(d=8\), \(c=0\), \(a=\frac{2}{125}\), and \(b=-\frac{6}{25}\). Thus, \(f(x)=\frac{2}{125}x^3-\frac{6}{25}x^2+8\). 3. Since \(f''(x)=\frac{12}{125}x-\frac{12}{25}\), the inflection point occurs at \(x=5\), where \(f(5)=4\). Thus, \(W=(5, 4)\). The slide is steepest there and changes concavity. 4. The slope at \(x=5\) is \(f'(5)=-\frac{6}{5}\). Therefore, the maximum downward grade is \(120\%\), and the angle of descent is \(\arctan(\frac{6}{5})\approx50.2^\circ\).

a) The slide reaches the ground at \(x=10\) and begins horizontally at \(x=0\). b) \(f(x)=\frac{2}{125}x^3-\frac{6}{25}x^2+8\) c) \(W=(5, 4)\); this is the steepest point and where concavity changes. d) \(120\%\); \(50.2^\circ\)

- Use the inflection point's coordinates to constrain the cubic. - What does a local extremum say about the first derivative? - Use the second derivative to determine the type of extremum. - For part c, keep the leading coefficient as a parameter.

1. Write \(f(x)=ax^3+bx^2+cx+d\). 2. The inflection point \(W=(0,1)\) gives \(d=1\) and \(f''(0)=2b=0\), so \(b=0\). 3. Since \(f'(-1)=0\), we have \(3a+c=0\), or \(c=-3a\). 4. Using \(P=(2,0)\), \(0=8a-6a+1=2a+1\). Thus \(a=-\frac{1}{2}\) and \(c=\frac{3}{2}\), giving \(f(x)=-\frac{1}{2}x^3+\frac{3}{2}x+1\). 5. Here \(f''(x)=-3x\), so \(f''(-1)=3>0\). The extremum is a local minimum. 6. In general the conditions at \(W\) and \(x=-1\) give \(f(x)=ax^3-3ax+1\). Passing through \(Q=(2,3)\) forces \(3=2a+1\), or \(a=1\). But then \(f''(-1)=-6<0\), so the extremum is a maximum. Diego's claim is true.

a) \(f(x)=-\frac{1}{2}x^3+\frac{3}{2}x+1\) b) Local minimum at \(x=-1\), since \(f''(-1)=3>0\). c) True. The point \(Q\) would force \(a=1\), giving \(f''(-1)=-6<0\) and a local maximum instead.

- What derivative behavior distinguishes a local minimum from a stationary inflection point? - What changes in the graph near an inflection point, and what does that mean for the derivative? - How much of a polynomial can its degree and zeros determine? - What additional information would fix a function's vertical position?

1. At the local minimum \(x=-\frac{3}{2}\), \(f'\) has a simple zero. At the stationary inflection point \(x=3\), \(f'\) has a double zero. Since \(f'\) is cubic, these are all its zeros. 2. The inflection-point x-coordinates of \(f\) are \(x=0\) and \(x=3\), so these are the zeros of \(f''\). Each is simple because the concavity changes at both values. 3. At \(x=0\), \(f''\) changes from positive to negative, so \(f'\) has a local maximum. At \(x=3\), \(f''\) changes from negative to positive, so \(f'\) has a local minimum. 4. The zeros and multiplicities give \(f'(x)=k\left(x+\frac{3}{2}\right)(x-3)^2\). The tangent slope at \(I\) gives \(f'(0)=2\), so \(2=k\left(\frac{3}{2}\right)(9)\), giving \(k=\frac{4}{27}\). Thus \(f'\) is uniquely determined. 5. If two polynomial functions have this same derivative, their difference has derivative zero everywhere. By the Mean Value Theorem, that difference is constant. Both must pass through \(I=(0,2)\), so the difference is zero. Therefore, \(f\) is uniquely determined without finding its formula.

a) Zeros of \(f'\): \(x=-\frac{3}{2}\), multiplicity \(1\); \(x=3\), multiplicity \(2\); zeros of \(f''\): \(x=0\) and \(x=3\), each with multiplicity \(1\). b) \(f'\) has a local maximum at \(x=0\) and a local minimum at \(x=3\). c) Yes. The zeros and \(f'(0)=2\) determine the derivative uniquely. Any two functions with that derivative differ by a constant (Mean Value Theorem), and \(f(0)=2\) fixes that constant.