- A turning point is where the graph changes from increasing to decreasing or from decreasing to increasing. - Trace the graph from left to right and count each direction change.

1. Reading from left to right, the graph changes direction three times: near \(x=-2\), \(x=0\), and \(x=2\). 2. At \(x=0\), the graph changes from increasing to decreasing, so that turning point is a local maximum.

The graph has \(3\) turning points. The turning point at \(x=0\) is a local maximum.

- Focus on how the direction changes at \(x=5\). - Compare “increasing then decreasing” with the definition of a local maximum or minimum.

1. The graph rises as it approaches \(x=5\). 2. It falls after \(x=5\), so the direction change is from increasing to decreasing. Therefore, the turning point is a local maximum.

A local maximum.

- Find where the curve stops falling and starts rising. - Read the horizontal and vertical coordinates at that point.

1. The graph decreases until \(x=1\) and increases after that point. 2. Its lowest point is \((1, -3)\), so this is a local minimum.

\((1, -3)\).

- A local maximum only needs to be higher than nearby points. - An absolute maximum must be at least as high as every point in the interval. - Compare the peak with every other turning point and with both endpoints.

1. On panel a), \(-1.5\le x\le1.5\), the peak at \(x=-1\) is higher than both endpoints and every other displayed point. It is an absolute maximum on that interval. 2. On panel b), \(-3\le x\le3\), the right endpoint reaches a greater value, so the peak at \(x=-1\) is only a local maximum on that interval.

a) Yes, an absolute maximum on the displayed interval. b) No, it is a local maximum but not an absolute maximum.

- A horizontal-looking part does not always mean the graph changes direction. - Compare whether each curve is rising or falling on the two sides of the origin.

1. In panel a), the curve rises before and after the origin. It flattens there but does not turn. 2. In panel b), the curve falls before the origin and rises afterward. It has a turning point, a local minimum, at the origin.

Graph b). Graph a) keeps rising; graph b) changes from falling to rising.

- Read the outputs from left to right and notice where their trend reverses. - A downward change from increasing to decreasing indicates which kind of extremum?

1. As \(x\) increases toward \(2\), the values rise from \(3\) to \(7\). 2. After \(x=2\), the values fall. Because the function is quadratic and the values are symmetric about \(x=2\), the turning point is a maximum at \((2, 7)\).

A maximum at \((2, 7)\).

- Use the vertex of the quadratic to find its minimum. - Determine what taking reciprocals does to the smallest and largest positive values. - Compare each function at \(-x\) and \(x\).

1. The function \(f\) has a global minimum of \(2\) at \(x=0\) and is unbounded above, so its range is \([2, \infty)\). 2. The reciprocal \(g\) has a global maximum of \(\frac{1}{2}\) at \(x=0\), is always positive, and approaches \(0\) without reaching it. Its range is \(\left(0, \frac{1}{2}\right]\). 3. Since \(f(-x)=f(x)\), \(f\) is even. Then \(g(-x)=\frac{1}{f(-x)}=\frac{1}{f(x)}=g(x)\), so \(g\) is also even. 4. Both extrema occur at \(x=0\), but \(f\) has a minimum at \((0, 2)\), while \(g\) has a maximum at \(\left(0, \frac{1}{2}\right)\).

1) Range of \(f\): \([2, \infty)\); range of \(g\): \(\left(0, \frac{1}{2}\right]\) 2) Both graphs are symmetric about the y-axis. 3) \(f\) has a minimum at \((0, 2)\); \(g\) has a maximum at \(\left(0, \frac{1}{2}\right)\).

- Start with familiar power functions. - Consider how the sign of the leading coefficient affects the graph. - For part c, look for a function that is increasing across its entire domain.

1. A quadratic with a positive leading coefficient opens upward and has one local minimum. One example is \(h(x)=x^2\). 2. The fourth-degree polynomial \(h(x)=-x^4\) has exactly one local maximum, at the origin. 3. A strictly increasing function has no local extrema. One example is \(h(x)=x^3\).

a) \(h(x)=x^2\) b) \(h(x)=-x^4\) c) \(h(x)=x^3\)

- Locate the two turning points first. - Trace the curve from left to right and note where y-values rise or fall as \(x\) increases. - Use the turning-point x-values as interval boundaries.

1. The graph rises until the local maximum at \(x=-1\). 2. It falls from \(x=-1\) to the local minimum at \(x=1\). 3. It rises again for \(x>1\).

Increasing on \((-\infty, -1)\) and \((1, \infty)\); decreasing on \((-1, 1)\).

- Start with the maximum-turning-points relationship for a degree-\(n\) polynomial. - The observed number of turning points cannot exceed that maximum. - Check the smallest degree for which the maximum is at least \(4\).

1. A degree-\(n\) polynomial can have at most \(n-1\) turning points. 2. Four turning points require \(n-1\ge4\), so \(n\ge5\). 3. A degree-\(5\) polynomial can attain the maximum of \(4\) turning points. Therefore, the least possible degree is \(5\).

The least possible degree is \(5\), because a degree-\(n\) polynomial has at most \(n-1\) turning points.

- Trace the curve from left to right rather than from bottom to top. - Identify each place where the curve switches direction. - Give one interval for each uninterrupted rise or fall.

1. Read from left to right. The curve falls until \(x=-2\), rises until \(x=0\), falls until \(x=2\), then rises again. 2. It is increasing on \((-2,0)\) and \((2,3)\). It is decreasing on \((-3,-2)\) and \((0,2)\).

Increasing: \((-2,0)\) and \((2,3)\). Decreasing: \((-3,-2)\) and \((0,2)\).

- What does an even function tell you about points on opposite sides of the y-axis? - How does replacing \(x\) with \(x-1\) move a graph? - Track the horizontal and vertical changes separately for each minimum.

1. Since \(f\) is even, its graph is symmetric about the y-axis. It also has a local minimum at \((-2,-3)\). 2. The rule \(g(x)=f(x-1)+2\) moves each point right \(1\) and up \(2\). 3. The two minima move to \((3,-1)\) and \((-1,-1)\).

\((-1,-1)\) and \((3,-1)\). Reflection across the y-axis gives the second minimum of \(f\); the translation moves both minima right \(1\) and up \(2\).

- A local maximum need not be the largest output on the whole domain. - Test a positive input farther to the right of the turning point. - What does the leading term tell you about the outputs for very large positive \(x\)?

1. The claim confuses a local maximum with an absolute maximum. At \(x=3\), \(f(3)=27-9+1=19\), which exceeds \(3\). 2. The positive cubic term dominates for large positive \(x\), so \(f(x)\) increases without bound as \(x\) increases. 3. Therefore, \(f\) has no absolute maximum on the real numbers.

For example, \(f(3)=19>3\). There is no absolute maximum because \(f(x)\) grows without bound as \(x\) increases.

- Identify the symmetry from the powers in the polynomial. - Reflect the positive x-coordinates across the y-axis. - Use the positive leading term and even degree to determine what happens on both ends. - Local extrema alternate in type along a smooth polynomial graph.

1. The polynomial contains only even powers, so it is even and its graph is symmetric about the y-axis. 2. Therefore, the extrema at \(x=1\) and \(x=2\) have matching extrema of the same type at \(x=-1\) and \(x=-2\). 3. The leading term is \(\frac{1}{6}x^6\), so \(f(x)\to\infty\) as \(x\to\pm\infty\). Thus, the outer extrema at \(x=\pm2\) are local minima. 4. The extrema alternate in type, so \(x=\pm1\) are local maxima and \(x=0\) is a local minimum.

a) \(x=-1\) and \(x=-2\) b) Local minima: \(x=-2, 0, 2\) Local maxima: \(x=-1, 1\)

- Identify the symmetry from the odd powers. - Reflect each extremum through the origin. - Determine what happens to a peak under a 180-degree rotation.

1. The polynomial contains only odd powers, so \(g\) is odd and its graph is symmetric about the origin. 2. The extrema at \(x=1\) and \(x=2\) are reflected to extrema at \(x=-1\) and \(x=-2\). 3. A 180-degree rotation about the origin changes a local maximum into a local minimum and a local minimum into a local maximum. 4. Therefore, \(x=-1\) is a local minimum, and \(x=-2\) is a local maximum.

a) \(x=-1\) and \(x=-2\) b) \(x=-1\) is a local minimum, and \(x=-2\) is a local maximum.

- Write the quadratic in vertex form. - A positive reciprocal is largest when its denominator is smallest. - What happens to \(\frac{1}{f(x)}\) when \(f(x)\) becomes very large?

1. Complete the square: \(f(x)=(x-3)^2+2\). Therefore, \(f\) has a global minimum at \((3, 2)\). 2. The reciprocal \(g(x)=\frac{1}{(x-3)^2+2}\) is largest when its positive denominator is smallest. Thus, \(g\) has a global maximum at \(\left(3, \frac{1}{2}\right)\). 3. As \(x\to\pm\infty\), \(f(x)\to\infty\), so \(g(x)\to 0\). The horizontal asymptote of \(g\) is \(y=0\). 4. More generally, for positive function values, taking reciprocals reverses order. A local minimum of \(f\) becomes a local maximum of \(\frac{1}{f}\), and a local maximum becomes a local minimum.

1) \(f\): global minimum at \((3, 2)\); \(g\): global maximum at \(\left(3, \frac{1}{2}\right)\) 2) \(f(x)\to\infty\) and \(g(x)\to 0\) as \(x\to\pm\infty\) 3) For positive \(f\), minima and maxima switch when taking the reciprocal.

- Local maxima and minima alternate. - Follow the only possible shape through the extrema from left to right. - Use the fact that the function is a polynomial and has no other extrema. - Count x-axis crossings from the end behavior and the positive extremum values.

1. Consecutive local extrema alternate between maxima and minima. Since the function value decreases from \(A\) to \(B\) and then increases from \(B\) to \(C\), \(A\) and \(C\) are local maxima, while \(B\) is a local minimum. 2. To the left of the leftmost local maximum, the polynomial must increase toward \(A\), so \(f(x)\to-\infty\) as \(x\to-\infty\). To the right of the rightmost local maximum, it must decrease away from \(C\), so \(f(x)\to-\infty\) as \(x\to\infty\). 3. All three extremum values are positive. The graph rises from \(-\infty\) to \(A\), so it crosses the x-axis exactly once before \(x=-2\). Between the extrema, the graph remains above the x-axis. After \(C\), it falls to \(-\infty\), so it crosses exactly once more. Therefore, \(f\) has exactly two real zeros.

a) \(A\) and \(C\) are local maxima; \(B\) is a local minimum. b) \(f(x)\to-\infty\) as \(x\to\pm\infty\). c) Exactly two real zeros.

- Substitute both given points into \(f(x) = ax^2 + bx + c\). - A downward-opening parabola has a negative leading coefficient. - Use the vertex formula \(x_v = -\frac{b}{2a}\) and require \(x_v > 0\).

1. Start with \(f(x) = ax^2 + bx + c\). Since the graph passes through \((0, 1.5)\), \(c = 1.5\). 2. Since the graph also passes through \((30, 0)\), \(900a + 30b + 1.5 = 0\). Solving for \(b\) gives \(b = -30a - 0.05\). Therefore, \(f_a(x) = ax^2 - (30a + 0.05)x + 1.5\). 3. The parabola opens downward when \(a < 0\). Its vertex has \(x\)-coordinate \(x_v = -\frac{b}{2a} = \frac{30a + 0.05}{2a}\). 4. Requiring \(x_v > 0\), while \(a < 0\), gives \(30a + 0.05 < 0\). Thus, \(a < -\frac{1}{600}\). 5. The maximum height occurs at the vertex, whose \(x\)-coordinate is \(x_v = \frac{30a + 0.05}{2a} = 15 + \frac{0.025}{a}\).

a) \(f_a(x) = ax^2 - (30a + 0.05)x + 1.5\) b) \(a < -\frac{1}{600}\) c) \(x = 15 + \frac{0.025}{a}\) feet from the nozzle

- Compare the two ends of an even-degree polynomial graph. - Think about what continuity forces when both ends rise or both ends fall. - Apply the stated degree-versus-turning-points relationship with \(n=4\).

1. For an even-degree polynomial, both ends rise when the leading coefficient is positive and both ends fall when the leading coefficient is negative. 2. Because polynomial graphs are continuous, a positive-leading even-degree polynomial must attain an absolute minimum, while a negative-leading even-degree polynomial must attain an absolute maximum. 3. A quartic has degree \(4\). Using the degree-turning-point bound, it can have at most \(4-1=3\) turning points.

a) A positive-leading even-degree polynomial has an absolute minimum; a negative-leading even-degree polynomial has an absolute maximum. b) A quartic can have at most \(3\) turning points because a degree-\(4\) polynomial can have at most \(4-1\) turning points.

- Identify each peak and valley. - Compare the y-values of all extrema. - Include the endpoint values when finding absolute extrema on a closed interval. - A local extremum need not be the highest or lowest point on the entire interval.

1. From the graph, the local maxima are \((-2.5, 3)\) and \((1, 4)\), and the local minima are \((-1, 1)\) and \((3, -3)\). 2. The endpoint values are \(f(-4)=-2\) and \(f(4)=-1\). The largest value on the interval is \(4\), so \((1, 4)\) is the absolute maximum. The smallest value is \(-3\), so \((3, -3)\) is the absolute minimum. The other two extrema are local only.

a) Local maxima: \((-2.5, 3)\), \((1, 4)\); local minima: \((-1, 1)\), \((3, -3)\) b) Absolute maximum: \((1, 4)\); absolute minimum: \((3, -3)\). The points \((-2.5, 3)\) and \((-1, 1)\) are local only.

- Use the sign of the leading term to determine end behavior. - Rewrite the function to identify its absolute extrema. - Compare the value at \(x=0\) and nearby values with the displayed graph.

1. The leading coefficient of \(k\) is negative and its degree is even, so \(k(x)\to-\infty\) as \(x\to\pm\infty\). The displayed graph rises at both ends. 2. Rewrite \(k(x)=2-(x^2-1)^2\). Therefore, \(k\) has absolute maxima of \(2\) at \(x=\pm1\), while the displayed graph has absolute minima. 3. Also, \(k(0)=1\), and nearby values such as \(k(0.5)=1.4375\) are larger, so \((0,1)\) is a local minimum. The displayed graph has a local maximum at \(x=0\) and y-intercept \(-1\).

1. \(k(x)\to-\infty\) as \(x\to\pm\infty\), but the displayed graph rises at both ends. 2. \(k\) has absolute maxima at \(x=\pm1\), while the graph has absolute minima. 3. \(k\) has a local minimum at \((0,1)\), while the graph has a local maximum at \((0,-1)\).

- Count the turning points and use the degree-turning-point bound. - Which powers can appear in a cubic with symmetry about the origin? - Read the coordinates of the two specified points and substitute into your general cubic rule. - Check the resulting equation against the remaining marked points.

1. The graph has two local extrema, so a polynomial must have degree at least \(3\). The least possible degree is \(3\). 2. The graph has origin symmetry, so under the cubic assumption its rule has the form \(f(x)=ax^3+cx\). 3. The marked point A at \((-1, 2)\) gives \(-a-c=2\). The marked point P at \((2, 2)\) gives \(8a+2c=2\). 4. From \(-a-c=2\), \(c=-2-a\). Substituting gives \(8a+2(-2-a)=2\), so \(a=1\) and \(c=-3\). 5. Thus, \(f(x)=x^3-3x\), which agrees with the other marked points and turning points.

1. Least possible degree: \(3\) 2. \(f(x)=ax^3+cx\) 3. \(f(x)=x^3-3x\)

- Relate the number of turning points to the least possible degree. - Which powers can appear in a fourth-degree polynomial with y-axis symmetry? - Read coordinates of the marked points, together with the additional function value. - Check the finished rule at both minima and at the central maximum.

1. The graph has three local extrema, so its degree must be at least \(4\). 2. It is symmetric about the y-axis. For an even polynomial of least degree \(4\), write \(f(x)=ax^4+bx^2+c\). 3. The marked maximum \((0,0)\) gives \(c=0\). The marked minimum \((2,-4)\) gives \(16a+4b=-4\), and the additional value \(f(1)=-\frac{7}{4}\) gives \(a+b=-\frac{7}{4}\). 4. Divide the first equation by \(4\) to get \(4a+b=-1\). Subtract \(a+b=-\frac{7}{4}\) to find \(3a=\frac{3}{4}\), so \(a=\frac{1}{4}\) and \(b=-2\). 5. Thus, \(f(x)=\frac{1}{4}x^4-2x^2\). Its marked minima at \(x=\pm2\) and maximum at \(x=0\) agree with the graph.

1. Least possible degree: \(4\) 2. Symmetric about the y-axis; \(f(x)=ax^4+bx^2+c\) 3. \(f(x)=\frac{1}{4}x^4-2x^2\)

- Interpret \(f(x)=c\) as intersections with a horizontal line \(y=c\). - Identify the y-values of the local maximum and the two global minima. - Check what happens when the horizontal line passes exactly through an extremum. - Consider the graph's end behavior beyond the visible turning points.

1. Solutions to \(f(x)=c\) are the x-coordinates where the horizontal line \(y=c\) intersects the graph. 2. For \(c=0\), the graph crosses the x-axis four times, so there are four solutions. 3. The graph has two global minima at height \(-4\) and a local maximum at height \(6\). A horizontal line intersects all four branches when \(-4<c<6\). 4. At \(c=-4\), the line is tangent at the two global minima, giving exactly two solutions. For \(c>6\), the line intersects only the two outer branches, also giving exactly two solutions. 5. At \(c=6\), the line passes through the local maximum and intersects both outer branches, giving exactly three solutions.

a) \(4\) solutions b) \(-4<c<6\) c) \(c=-4\) or \(c>6\) d) \(c=6\)

- What two quantities must you multiply to calculate ticket revenue? - How can you rewrite a quadratic with a negative coefficient of the squared term to see its highest value? - After finding the ticket count, use the original price rule to find the price.

1. Revenue equals tickets times price: \(R(n)=n(30-0.2n)=-0.2n^2+30n\). 2. Complete the square: \(R(n)=-0.2(n-75)^2+1125\). The squared term is nonnegative, so the greatest revenue occurs at \(n=75\) and equals \(\$1125\). 3. The price is \(p(75)=30-0.2(75)=\$15\) per ticket.

a) \(R(n)=-0.2n^2+30n\). b) \(75\) tickets; greatest revenue \(\$1125\). c) \(\$15\) per ticket.

- Interpret \(g(x)=k\) as intersections with a horizontal line \(y=k\). - Track what changes when the line passes through the minimum, the x-axis, and the maximum. - Remember that the right branch approaches \(y=0\) but does not reach it again.

1. Zeros are x-intercepts. The graph crosses the x-axis at \(x=-3\) and \(x=0\), so \(g\) has two zeros. 2. A horizontal line \(y=k\) intersects the graph twice when \(-2<k\le0\). It also intersects twice when \(k=2\): once on the left branch and once at the local maximum. 3. When \(0<k<2\), the horizontal line intersects the left branch once and the right-hand hump twice, for three solutions. 4. When \(k<-2\), the horizontal line lies below the global minimum and does not intersect the graph. At \(k=-2\), there is one solution.

a) \(2\) b) \(k\in(-2, 0]\cup\{2\}\) c) \(k\in(0, 2)\) d) \(k<-2\)