- Build the complete candidate list before comparing values. - Endpoints and interior critical numbers play the same role in the final comparison. - Check whether the largest or smallest value occurs more than once.

1. The candidates are the endpoints \(x=-3,5\) and the interior critical numbers \(x=-1,4\). 2. The largest candidate value is \(6\), attained at \(x=-3\) and \(x=4\). 3. The smallest candidate value is \(-2\), attained at \(x=-1\).

Absolute maxima: \(f(-3)=f(4)=6\). Absolute minimum: \(f(-1)=-2\).

- Find the beginning and end of the closed interval. - The candidates test uses more than just points where the derivative is zero.

1. The closed domain begins at \(x=0\) and ends at \(x=3\). 2. The candidates test always includes both endpoints, regardless of where the derivative is zero.

\(x=0\) and \(x=3\).

- Which quantity are you ultimately trying to make largest or smallest? - What is the job of the derivative in finding candidates? - Which number gives the actual height of the curve at \(2\)?

1. The condition \(f'(2)=0\) identifies \(x=2\) as a critical number. 2. Absolute extrema concern the original function's heights, so the value to compare is \(f(2)\), not its derivative.

Compare \(f(2)\). The derivative identifies the candidate, but the function value determines the extrema.

- Include both ends of the closed interval. - Can a corner be a critical number even without a horizontal tangent? - Combine the endpoints with every interior critical number.

1. The two endpoints, \(x=0\) and \(x=3\), are candidates. 2. At the interior corner \(x=1\), the derivative does not exist while \(f\) is defined, so \(1\) is also critical. 3. The complete candidate set is \(\{0,1,3\}\).

\(x=0\), \(x=1\), and \(x=3\).

- Check that all candidates have already been listed. - Compare the function values, not the x-coordinates. - A minimum might be inside the interval while a maximum occurs at an endpoint.

1. The table includes both endpoints and the only interior critical number, so all candidates are listed. 2. The greatest value is \(5\) at \(x=3\). 3. The least value is \(2\) at \(x=1\).

Absolute maximum \(5\) at \(x=3\); absolute minimum \(2\) at \(x=1\).

- Look for all places that reach the greatest height, not just one. - Can an absolute maximum occur at two different inputs? - Evaluate the function at the two highest points.

1. The graph reaches its highest height at \(x=-1\) and \(x=1\). 2. At both points, \(f(x)=-(1-1)^2=0\). 3. No point of the curve lies above the x-axis, so both points attain the absolute maximum.

Absolute maximum value \(0\), attained at \(x=-1\) and \(x=1\).

- Turning points separate increasing and decreasing intervals. - A critical number can occur where the derivative is zero or where the derivative does not exist, provided the function itself is defined. - Look for interior corners rather than smooth turning points in this graph. - For the candidates test, include both endpoints and all interior critical numbers before comparing values.

1. The water level increases on \([0,3]\), \([6,9]\), and \([12,13]\), and decreases on \([3,6]\) and \([9,12]\). 2. The interior local maxima are \((3,5)\) and \((9,4)\). The interior local minima are \((6,0.5)\) and \((12,1)\). 3. At \(t=3,6,9,12\), two line segments meet with different slopes, so \(h'\) does not exist. Because \(h\) is defined at those inputs, all four are critical numbers. 4. Compare \(h(0)=2\), \(h(3)=5\), \(h(6)=0.5\), \(h(9)=4\), \(h(12)=1\), and \(h(13)=1.5\). The absolute minimum is \(0.5\,\text{m}\) at \(t=6\,\text{h}\).

a) Increasing on \([0,3]\), \([6,9]\), and \([12,13]\); decreasing on \([3,6]\) and \([9,12]\). b) Interior local maxima: \((3,5)\), \((9,4)\); interior local minima: \((6,0.5)\), \((12,1)\). c) The interior vertices \(t=3,6,9,12\) are critical numbers because \(h\) is defined there but \(h'\) is not. Candidate values: \((0,2),(3,5),(6,0.5),(9,4),(12,1),(13,1.5)\). The absolute minimum is \(0.5\,\text{m}\) at \(t=6\,\text{h}\).

- Determine whether the derivative can equal zero inside the interval. - A closed interval always contributes its endpoints to the candidate set. - Compare the endpoint values after confirming there are no other candidates.

1. \(f'(x)=-e^{-x}<0\) for every real \(x\), so there are no interior critical numbers. 2. The candidates test still includes the endpoints. \(f(-2)=e^2\) and \(f(1)=e^{-1}\). 3. Therefore, the absolute maximum is \(e^2\) at \(x=-2\), and the absolute minimum is \(e^{-1}\) at \(x=1\).

Absolute maximum: \(e^2\) at \(x=-2\). Absolute minimum: \(e^{-1}\) at \(x=1\). Endpoints remain candidates even when there are no interior critical numbers.

- Critical numbers include more than zeros of the derivative. - Make the candidate set from the interval boundaries and all interior critical numbers. - The values are already supplied, so the final step is a comparison.

1. The candidates are the endpoints \(-2\) and \(6\), the point \(x=1\) where the derivative is undefined, and the derivative zero \(x=4\). 2. Comparing the given values, the largest is \(7\) at \(x=1\), and the smallest is \(-2\) at \(x=4\).

Absolute maximum: \(7\) at \(x=1\). Absolute minimum: \(-2\) at \(x=4\).

- Distinguish a conclusion about nearby points from a conclusion about the entire interval. - Ask which locations a closed-interval absolute-extrema test requires. - A local classification does not replace a value comparison.

1. The condition \(f''(2)<0\) at a critical number establishes a local maximum, not necessarily an absolute maximum on the whole interval. 2. To determine an absolute maximum on \([0,5]\), evaluate \(f\) at both endpoints and at every interior critical number. 3. The largest of all candidate values is the absolute maximum.

The second derivative test gives only a local classification. The candidates test must compare \(f(0)\), \(f(5)\), and \(f\) at every interior critical number before an absolute maximum can be identified.

- For a continuous function on a closed interval, include both endpoints and every interior critical number. - A corner is a critical number when the function is defined there but the derivative does not exist. - Do not choose the absolute extrema until every candidate value has been entered in the table.

1. The endpoints are \(x=-5\) and \(x=3\). The interior polyline corners are \(x=-3,0,2\). At each corner, \(f\) is defined but the left and right slopes differ, so \(f'\) does not exist; therefore each is a critical number. 2. The complete candidate set is \(\{-5,-3,0,2,3\}\). 3. Candidate table: \(f(-5)=4\), \(f(-3)=-2\), \(f(0)=3\), \(f(2)=-2\), \(f(3)=1\). 4. Comparing the candidate values gives the absolute maximum \(4\) at \(x=-5\), and the absolute minimum \(-2\) at both \(x=-3\) and \(x=2\).

Candidate set: \(\{-5,-3,0,2,3\}\). Candidate values: \(4,-2,3,-2,1\), respectively. Absolute maximum: \(4\) at \(x=-5\). Absolute minima: \(-2\) at \(x=-3\) and \(x=2\).

- What theorem guarantees that the extrema exist? - Where could an absolute extremum occur inside a differentiable part of the interval? - Have endpoints and missing-derivative points both been included?

1. Continuity on a closed interval guarantees absolute extrema exist. 2. An absolute extremum is either at an endpoint or at an interior critical number. 3. The two endpoints are \(-2\) and \(4\), and the only interior critical numbers are \(-1\) and \(1\). 4. Therefore no other inputs can produce an absolute extremum that is not among these candidates.

The listed points are both endpoints and every interior critical number. By the Extreme Value Theorem and the critical-point condition, comparing their function values is exhaustive.

- Finding a zero derivative does not finish the candidates test. - Which extra x-values must you evaluate on a closed interval? - Compare the original function values at every candidate.

1. The derivative \(f'(x)=2x\) vanishes at the interior critical number \(0\). 2. The candidates test also includes the endpoints \(-1\) and \(2\). 3. Their values are \(f(-1)=1\), \(f(0)=0\), and \(f(2)=4\). 4. The absolute minimum is \(0\) at \(0\), and the absolute maximum is \(4\) at \(2\).

Mateo omitted both endpoints. The absolute minimum is \(0\) at \(x=0\); the absolute maximum is \(4\) at \(x=2\).

- The maximum has to beat both other candidate values. - What happens if another candidate equals \(7\)? - Turn “the only location” into a strict comparison.

1. For a unique absolute maximum at \(x=2\), the value \(7\) must be strictly greater than every other candidate value. 2. It is already greater than \(3\). 3. The remaining requirement is \(7>k\), or \(k<7\).

\(k<7\)

- What features of the given interval ensure that absolute extrema are attained? - What are the possible locations of those extrema? - How can you demonstrate that no candidate has been omitted before making the comparison?

1. Differentiate: \(f'(x)=-2x+4\), which is zero at the interior critical number \(x=2\). 2. The endpoint and critical-number candidates are \(x=0,2,5\). Their values are \(f(0)=1\), \(f(2)=5\), and \(f(5)=-4\). 3. The function is continuous on the closed interval. Comparing all candidate values gives the absolute maximum \(5\) at \(x=2\) and the absolute minimum \(-4\) at \(x=5\).

Candidates: \((0,1)\), \((2,5)\), and \((5,-4)\). Absolute maximum: \(5\) at \(x=2\). Absolute minimum: \(-4\) at \(x=5\).

- Which interior inputs could possibly attain an absolute extremum? - How does including both endpoints affect the candidate list? - What happens if multiple candidates have equal outputs?

1. Differentiate: \(f'(x)=3x^2-3\); the critical numbers are \(-1\) and \(1\). 2. The endpoint and critical values are \(f(-2)=-2\), \(f(-1)=2\), \(f(1)=-2\), and \(f(2)=2\). 3. The absolute maximum is \(2\), attained at \(-1\) and \(2\); the absolute minimum is \(-2\), attained at \(-2\) and \(1\).

Candidates: \(f(-2)=-2\), \(f(-1)=2\), \(f(1)=-2\), and \(f(2)=2\). Absolute maximum: \(2\) at \(x=-1,2\). Absolute minimum: \(-2\) at \(x=-2,1\).

- Check which endpoints belong to the domain. - A value approached but never reached is not an absolute maximum. - Determine whether any allowed input gives the smallest value.

1. The function increases throughout its domain. 2. Its least value is \(0\), reached at the included endpoint \(x=0\). 3. Its values approach \(1\), but \(x=1\) is excluded, so it has no absolute maximum. The interval is not closed, so the closed-interval candidates test does not apply.

Absolute minimum \(0\) at \(x=0\). No absolute maximum exists. The closed-interval candidates test is inapplicable because the right endpoint is excluded.

- What kinds of inputs should be compared for an absolute-extremum problem on a closed interval? - Which features of the temperature model allow you to find all candidates? - How will you account for more than one time attaining the same extreme value?

1. Differentiate to get \(T'(t)=3t^2-12t+9=3(t-1)(t-3)\), so the interior candidates are \(t=1\) and \(t=3\). 2. The candidate temperatures are \(T(0)=15\), \(T(1)=19\), \(T(3)=15\), and \(T(5)=35\). 3. The highest is \(35\,{}^\circ\text{F}\) at \(t=5\,\text{h}\); the lowest is \(15\,{}^\circ\text{F}\) at \(t=0\,\text{h}\) and \(t=3\,\text{h}\).

Candidates: \(T(0)=15\), \(T(1)=19\), \(T(3)=15\), and \(T(5)=35\), in degrees Fahrenheit. Highest: \(35\,{}^\circ\text{F}\) at \(5\,\text{h}\). Lowest: \(15\,{}^\circ\text{F}\) at both \(0\,\text{h}\) and \(3\,\text{h}\).

- First find the critical numbers in the interior of the interval. - Use the second derivative to classify the interior critical points. - Include both endpoints when looking for absolute extrema. - Compare all candidate function values.

1. Differentiate: \(f'(x) = x^3 - 4x = x(x - 2)(x + 2)\). The critical numbers are \(x = -2\), \(x = 0\), and \(x = 2\). 2. Use \(f''(x) = 3x^2 - 4\). Since \(f''(-2) = 8 > 0\) and \(f''(2) = 8 > 0\), \((-2, -1)\) and \((2, -1)\) are local minima. Since \(f''(0) = -4 < 0\), \((0, 3)\) is a local maximum. 3. Evaluate the endpoints and critical numbers: \(f(-3) = \frac{21}{4} = 5.25\), \(f(-2) = -1\), \(f(0) = 3\), \(f(2) = -1\), and \(f(2.5) = \frac{17}{64} = 0.265625\). 4. Comparing these values, the absolute maximum is \((-3, 5.25)\), and the absolute minima are \((-2, -1)\) and \((2, -1)\).

Local maximum: \((0, 3)\) Local minima: \((-2, -1)\) and \((2, -1)\) Absolute maximum: \((-3, 5.25)\) Absolute minima: \((-2, -1)\) and \((2, -1)\)

- Find the critical numbers by setting the first derivative equal to zero. - Use the second derivative to classify the critical points. - Why must the endpoint values also be checked on a closed interval? - An absolute extremum is the highest or lowest value on the entire interval.

1. Differentiate: \(g'(x) = \frac{1}{2}x^2 - x - \frac{3}{2}\). 2. Solve \(g'(x) = 0\): \(x^2 - 2x - 3 = 0\), so \(x = -1\) and \(x = 3\). 3. Use \(g''(x) = x - 1\). Since \(g''(-1) = -2 < 0\), \((-1, \frac{17}{6})\) is a local maximum. Since \(g''(3) = 2 > 0\), \((3, -\frac{5}{2})\) is a local minimum. 4. Evaluate the endpoints: \(g(-2) = \frac{5}{3}\) and \(g(4) = -\frac{4}{3}\). 5. Compare the values. The local maximum at \((-1, \frac{17}{6})\) is also the absolute maximum, and the local minimum at \((3, -\frac{5}{2})\) is also the absolute minimum.

Local and absolute maximum: \((-1, \frac{17}{6})\) Local and absolute minimum: \((3, -\frac{5}{2})\)

- First simplify the sum \(f(x) + g(x)\). - Find critical numbers in the interior by setting the derivative equal to zero. - Do not forget to evaluate the function at both endpoints. - Compare all candidate values to identify the absolute extrema.

1. Combine the functions: \(s(x) = x^2 - 4x + 6 - 0.5x^2 + x + 2 = 0.5x^2 - 3x + 8\). 2. Differentiate: \(s'(x) = x - 3\). The only interior critical number is \(x = 3\). 3. Evaluate \(s\) at the critical number and endpoints: \(s(0) = 8\), \(s(3) = 3.5\), and \(s(6) = 8\). 4. The absolute minimum is \(3.5\) at the interior point \(x = 3\). The absolute maximum is \(8\) at the endpoints \(x = 0\) and \(x = 6\).

Absolute minimum: \(s(3) = 3.5\), attained at the interior point \(x = 3\) Absolute maximum: \(s(0) = s(6) = 8\), attained at the endpoints \(x = 0\) and \(x = 6\)

- Distribute the subtraction sign across the entire expression for \(g(x)\). - Which points must be checked for absolute extrema on a closed interval? - Keep only critical numbers that lie in the stated interval. - Compare the values at every candidate point.

1. Simplify the difference: \(d(x) = \frac{1}{3}x^3 - x^2 + 4 - (-x^2 + x + 1) = \frac{1}{3}x^3 - x + 3\). 2. Differentiate: \(d'(x) = x^2 - 1\). Solving \(d'(x) = 0\) gives \(x = -1\) and \(x = 1\). Only \(x = 1\) lies in \([0, 3]\). 3. Evaluate \(d\) at the interior critical number and endpoints: \(d(0) = 3\), \(d(1) = \frac{7}{3}\), and \(d(3) = 9\). 4. The absolute minimum is \(\frac{7}{3}\) at the interior point \(x = 1\). The absolute maximum is \(9\) at the endpoint \(x = 3\).

Absolute minimum: \(d(1) = \frac{7}{3}\), attained at the interior point \(x = 1\) Absolute maximum: \(d(3) = 9\), attained at the endpoint \(x = 3\)

- Which input must be excluded from the domain? - How can the derivative behavior distinguish a minimum from a maximum when the parameter changes sign? - What must be true of the function and interval before the Extreme Value Theorem applies? - How could you organize all possible absolute-extremum locations before comparing their function values?

1. Since \(x^2\ne0\), the domain is \(\mathbb{R}\setminus\{0\}\). 2. Differentiate: \(f_k'(x)=\frac{2(x-k)}{x^3}\). The only critical number in the domain is \(x=k\), and \(f_k(k)=-\frac{1}{k}\). 3. If \(k>0\), the derivative changes from negative to positive at \(k\), giving a local minimum. If \(k<0\), it changes from positive to negative, giving a local maximum. 4. For \(k>0\), the function is continuous on the closed interval \(\left[\frac{k}{2},2k\right]\); the Extreme Value Theorem guarantees absolute extrema there. The candidates are the two endpoints and the interior critical number \(k\). 5. Their values are \(f_k\left(\frac{k}{2}\right)=0\), \(f_k(k)=-\frac{1}{k}\), and \(f_k(2k)=-\frac{3}{4k}\). Since \(k>0\), \(-\frac{1}{k}<-\frac{3}{4k}<0\). The absolute maximum is \(0\) at \(x=\frac{k}{2}\), and the absolute minimum is \(-\frac{1}{k}\) at \(x=k\).

a) Domain: \(\mathbb{R}\setminus\{0\}\). Local extremum: \(\left(k,-\frac{1}{k}\right)\), a minimum for \(k>0\) and a maximum for \(k<0\). b) Candidates: \(\left(\frac{k}{2},0\right)\), \(\left(k,-\frac{1}{k}\right)\), and \(\left(2k,-\frac{3}{4k}\right)\). Absolute maximum: \(0\) at \(x=\frac{k}{2}\). Absolute minimum: \(-\frac{1}{k}\) at \(x=k\).

- How could you ensure that the list of possible absolute-extremum inputs is complete? - What evidence makes an interior input worth considering? - How can you distinguish an absolute winner from a value that is only locally largest? - What further information would describe how the output changes between candidates?

1. Differentiate: \(P'(x)=-6x^2+48x=-6x(x-8)\). The interior critical number in \((0,12)\) is \(x=8\). 2. The closed-interval candidate set is \(\{0,8,12\}\). 3. Compare candidate values: \(P(0)=0\), \(P(8)=512\), and \(P(12)=0\). Therefore, the absolute maximum is \(512\,\text{kW}\) at \(x=8\), corresponding to \(8000\) revolutions per minute. The absolute minimum power is \(0\,\text{kW}\), attained at both endpoints. 4. Since \(P'(x)>0\) on \((0,8)\) and \(P'(x)<0\) on \((8,12)\), the power is increasing on \([0,8]\) and decreasing on \([8,12]\).

Candidate set: \(\{0,8,12\}\). Candidate values: \(P(0)=0\), \(P(8)=512\), \(P(12)=0\). The absolute maximum power is \(512\,\text{kW}\) at \(8000\) revolutions per minute. Increasing on \([0,8]\); decreasing on \([8,12]\).

- The candidates test starts with endpoints plus every interior critical number. - A visible turning point is an interior critical-number candidate here. - Do not choose the highest or lowest point by inspection alone; write the candidate values first and compare that finite list.

1. The endpoints are \(x=-4\) and \(x=2\). The interior turning points occur at \(x=-3,-1,1\), so the candidate set is \(\{-4,-3,-1,1,2\}\). 2. Reading the graph gives \(f(-4)=4.25\), \(f(-3)=-2\), \(f(-1)=2\), \(f(1)=-2\), and \(f(2)=4.25\). 3. Comparing only these candidate values, the absolute maximum value is \(4.25\), attained at \(x=-4\) and \(x=2\). The absolute minimum value is \(-2\), attained at \(x=-3\) and \(x=1\). 4. The interior candidates at \(x=-3\) and \(x=1\) are local minima; \(x=-1\) is a local maximum.

Candidate table: \((-4,4.25),(-3,-2),(-1,2),(1,-2),(2,4.25)\). Absolute maxima: \(x=-4,2\), value \(4.25\). Absolute minima: \(x=-3,1\), value \(-2\). Interior local minima: \(x=-3,1\); interior local maximum: \(x=-1\).

- Find every interior critical number from \(T'(t)=0\), then add both endpoints. - Evaluate \(T\) at every candidate before choosing the absolute maximum. - Convert hours after 6:00 a.m. to clock time after the maximizing t-value is known. - Use the sign of \(T'\) only after completing the candidate comparison.

1. Differentiate: \(T'(t)=-0.15t^2+1.2t=-0.15t(t-8)\). The interior critical number is \(t=8\). 2. The complete closed-interval candidate set is \(\{0,8,12\}\). 3. Compare candidate values: \(T(0)=5\), \(T(8)=17.8\), and \(T(12)=5\). Therefore, the absolute maximum is \(17.8^\circ\text{C}\) at \(t=8\). 4. Eight hours after 6:00 a.m. is 2:00 p.m. 5. Since \(T'(t)>0\) on \((0,8)\) and \(T'(t)<0\) on \((8,12)\), the temperature increases from 6:00 a.m. to 2:00 p.m. and decreases from 2:00 p.m. to 6:00 p.m.

Candidate set: \(\{0,8,12\}\). Candidate values: \(T(0)=5\), \(T(8)=17.8\), \(T(12)=5\). The absolute maximum temperature is \(17.8^\circ\text{C}\) at 2:00 p.m. The temperature increases from 6:00 a.m. to 2:00 p.m. and decreases from 2:00 p.m. to 6:00 p.m.

- Find critical times of the velocity and compare them with the interval endpoints. - Acceleration is the derivative of velocity. - The most rapid decrease occurs where acceleration is smallest. - Compare any acceleration critical point with the interval endpoints if needed.

1. Differentiate: \(v'(t)=0.03t^2-1.2t+9\). The critical times are \(t=10\) and \(t=30\). Compare the velocity at the endpoints and the interior critical time: \(v(0)=0\), \(v(10)=40\), and \(v(30)=0\). Therefore, the maximum velocity is \(40\,\text{m/s}\) at \(t=10\,\text{s}\). 2. Acceleration is \(a(t)=v'(t)\). Its derivative is \(a'(t)=v''(t)=0.06t-1.2\), which is zero at \(t=20\). Since acceleration is an upward-opening quadratic, this is its minimum. 3. The acceleration is \(a(20)=-3\,\text{m/s}^2\), so the velocity is decreasing most rapidly then; the deceleration magnitude is \(3\,\text{m/s}^2\).

a) \(40\,\text{m/s}\) at \(t=10\,\text{s}\). b) At \(t=20\,\text{s}\), the acceleration is \(-3\,\text{m/s}^2\), corresponding to a deceleration magnitude of \(3\,\text{m/s}^2\).

- What would make a list of possible absolute-extremum inputs complete? - How could algebra confirm that every turning point suggested by the graph has been considered? - How can a table help you check for ties among the candidate values? - What evidence, besides a visual impression, establishes an interval of decrease?

1. Differentiate: \(f'(x)=0.4x(x^2-4)\). Its zeros in the interval are \(x=-2,0,2\), and the endpoints are \(x=-3,3\). These are all the candidates. 2. Substitute into the formula to obtain the exact values \(f(-3)=1.9\), \(f(-2)=-0.6\), \(f(0)=1\), \(f(2)=-0.6\), and \(f(3)=1.9\). 3. The function is continuous on the closed interval, and these are all the endpoints and interior critical numbers. Comparing their values shows that the largest value is \(1.9\) at both endpoints and the smallest is \(-0.6\) at \(x=-2,2\). 4. The derivative is negative on \((-3,-2)\) and \((0,2)\), so \(f\) is strictly decreasing on \([-3,-2]\) and \([0,2]\), consistent with the graph.

a) \(f'(x)=0.4x(x^2-4)\), so the interior critical numbers are \(-2\), \(0\), and \(2\). <table><tr><th>Candidate \(x\)</th><th>\(f(x)\)</th></tr><tr><td>\(-3\)</td><td>\(1.9\)</td></tr><tr><td>\(-2\)</td><td>\(-0.6\)</td></tr><tr><td>\(0\)</td><td>\(1\)</td></tr><tr><td>\(2\)</td><td>\(-0.6\)</td></tr><tr><td>\(3\)</td><td>\(1.9\)</td></tr></table> b) These are all endpoints and interior critical numbers, so no other inputs can give absolute extrema. Absolute maxima: \(1.9\) at \(x=-3,3\). Absolute minima: \(-0.6\) at \(x=-2,2\). c) Strictly decreasing on \([-3,-2]\) and \([0,2]\).

- Include both endpoints and all interior critical numbers as candidates. - Compare the function values at every candidate. - Fermat's theorem concerns differentiable interior extrema, not endpoints.

1. Differentiate: \(g'(x)=0.3x^2-1.2x+0.9=0.3(x-1)(x-3)\). The critical numbers in \([0, 5]\) are \(x=1\) and \(x=3\). 2. Evaluate \(g\) at the endpoints and critical numbers: \(g(0)=2\), \(g(1)=2.4\), \(g(3)=2\), and \(g(5)=4\). The greatest value is \(4\), so the absolute maximum is \(g(5)=4\). 3. The maximum occurs at the endpoint \(x_e=5\). Fermat's theorem applies only to differentiable functions at interior local extrema, so it does not require an endpoint derivative to be zero. In fact, \(g'(5)=2.4\ne0\). 4. The condition \(g'(x_0)=0\) is necessary when \(x_0\) is an interior point of the domain, \(g\) is differentiable at \(x_0\), and \(g\) has a local extremum there.

1) Absolute maximum: \(g(5)=4\) 2) The maximum occurs at the endpoint \(x=5\), so the derivative need not be zero. 3) The zero-derivative condition is necessary for a differentiable function at an interior local extremum.

- The candidate set is not just the turning points: include both endpoints. - Write the value of \(f\) at every candidate before classifying any absolute extremum. - Local classification uses nearby behavior; absolute classification comes from comparing the candidate values.

1. The endpoints are \(x=-3\) and \(x=3\). The interior turning points are \(x=-2,0,2\). Therefore the candidate set is \(\{-3,-2,0,2,3\}\). 2. The graph gives \(f(-3)=6.25\), \(f(-2)=0\), \(f(0)=4\), \(f(2)=0\), and \(f(3)=6.25\). 3. Comparing the candidate values, the absolute maxima occur at \(x=-3\) and \(x=3\), and the absolute minima occur at \(x=-2\) and \(x=2\). 4. The points \(x=-2\) and \(x=2\) are local minima. The point \(x=0\) is a local maximum.

Candidates: \((-3,6.25),(-2,0),(0,4),(2,0),(3,6.25)\). Absolute maxima: \(x=-3,3\). Absolute and local minima: \(x=-2,2\). Local maximum: \(x=0\).

- On a closed interval, the candidates test includes interior critical points and both endpoints. - Find where the first derivative is zero inside the interval. - Compare the function values at every candidate rather than using derivative sign alone.

1. Differentiate: \(f'(x)=\frac{1}{2}-\frac{2}{x^2}\). 2. A critical point occurs when \(f'(x)=0\). Solving \(\frac{1}{2}-\frac{2}{x^2}=0\) gives \(x^2=4\), so the only critical point in \((1,6)\) is \(x=2\). 3. Evaluate the candidates: \(f(1)=\frac{5}{2}\), \(f(2)=2\), and \(f(6)=\frac{10}{3}\). 4. The smallest candidate value is \(2\), so the absolute minimum is \(2\) at \(x=2\). The largest is \(\frac{10}{3}\), so the absolute maximum is \(\frac{10}{3}\) at \(x=6\).

Critical point: \(x=2\). Absolute minimum: \(2\) at \(x=2\). Absolute maximum: \(\frac{10}{3}\) at \(x=6\).

- Use the derivative of velocity to locate candidates for maximum velocity. - Compare interior candidates with the interval endpoints. - Acceleration is the derivative of velocity. - Analyze the acceleration function to find its maximum.

1. \(v'(t)=-0.15t^2+1.2t=-0.15t(t-8)\). Comparing critical points and endpoints on \([0,12]\) shows that the maximum occurs at \(t=8\), where \(v(8)=12.8\,\text{m/s}\). 2. The acceleration is \(a(t)=-0.15t^2+1.2t\). Since \(a\) is a downward-opening parabola, its maximum occurs at \(t=4\), where \(a(4)=2.4\,\text{m/s}^2\).

a) At \(t=8\,\text{s}\); \(12.8\,\text{m/s}\) b) At \(t=4\,\text{s}\); \(2.4\,\text{m/s}^2\)

- A critical number can occur where the derivative is zero or where it fails to exist. - Check the point where the absolute-value expression changes form. - Compare the function values only after the full candidate set is identified.

1. The function is continuous on \([-2,5]\) and is not differentiable at \(x=1\), so \(x=1\) is an interior critical number. 2. For \(x<1\), \(f'(x)=-1+\frac13=-\frac23\). For \(x>1\), \(f'(x)=1+\frac13=\frac43\). There are no other interior critical numbers. 3. Evaluate the candidates: \(f(-2)=3-\frac23=\frac73\), \(f(1)=\frac13\), and \(f(5)=4+\frac53=\frac{17}{3}\). 4. Therefore, the absolute minimum is \(\frac13\) at \(x=1\), and the absolute maximum is \(\frac{17}{3}\) at \(x=5\).

Absolute minimum: \(\frac13\) at \(x=1\). Absolute maximum: \(\frac{17}{3}\) at \(x=5\).

- Use the derivative graph to find interior candidate locations. - Include both endpoints when building the full candidate list. - Compare the listed function values to identify the absolute extrema.

1. The graph of \(f'\) has zeros at \(x=-2,1,3\). These are the interior critical numbers. 2. Together with the endpoints, the candidate set is \(\{-4,-2,1,3,4\}\). 3. The greatest candidate value in the table is \(f(-2)=4\). The smallest is \(f(1)=1\). 4. Therefore, the absolute maximum is \(4\) at \(x=-2\), and the absolute minimum is \(1\) at \(x=1\).

Interior critical numbers: \(x=-2,1,3\). Absolute maximum: \(4\) at \(x=-2\). Absolute minimum: \(1\) at \(x=1\).

- Ask whether interior critical numbers are the entire candidate list on a closed interval. - Evaluate the original function at every required location. - Watch for ties when identifying absolute extrema.

1. On a closed interval, the candidates test requires checking both endpoints in addition to every interior critical number. 2. Evaluate all candidates: \(f(-1)=-2\), \(f(0)=2\), \(f(2)=-2\), and \(f(3)=2\). 3. The absolute maximum value is \(2\), attained at \(x=0\) and \(x=3\). The absolute minimum value is \(-2\), attained at \(x=-1\) and \(x=2\).

a) Ieva omitted the endpoints. b) Absolute maxima: \(2\) at \(x=0\) and \(x=3\). Absolute minima: \(-2\) at \(x=-1\) and \(x=2\).

- Find every interior location where the derivative can make a candidate. - Add both endpoints before evaluating the original function. - The same absolute value may occur at more than one type of candidate.

1. \(f'(x)=3x^2-12x+9=3(x-1)(x-3)\), so the interior critical numbers are \(x=1\) and \(x=3\). 2. Evaluate the candidates: \(f(0)=4\), \(f(1)=8\), \(f(3)=4\), and \(f(5)=24\). 3. The absolute maximum is \(24\) at the endpoint \(x=5\). The absolute minimum is \(4\), attained at the endpoint \(x=0\) and the interior critical number \(x=3\).

Absolute maximum: \(24\) at the endpoint \(x=5\). Absolute minimum: \(4\) at \(x=0\) and \(x=3\).

- Find the interior candidate by setting the derivative equal to zero. - Use a right-triangle relationship to evaluate the trigonometric expression exactly at that angle. - Do not compare the interior value until the two endpoint values are included.

1. \(f'(x)=2\cos x-\sin x\). Critical numbers satisfy \(\tan x=2\). There is one such value \(c\in(0,\pi/2)\). 2. For this angle, \(\sin c=\frac{2}{\sqrt5}\) and \(\cos c=\frac{1}{\sqrt5}\), so \(f(c)=\sqrt5\). 3. The endpoint values are \(f(0)=1\) and \(f(\pi)=-1\). 4. Comparing candidates gives the absolute maximum \(\sqrt5\) at \(x=\arctan 2\), and the absolute minimum \(-1\) at \(x=\pi\).

Absolute maximum: \(\sqrt5\) at \(x=\arctan 2\). Absolute minimum: \(-1\) at \(x=\pi\).

- Recall the relationship between velocity and acceleration. - An interior maximum of acceleration can occur where the derivative of acceleration is zero. - Compare every eligible critical point with the endpoints of the test interval. - Include acceleration units in the final answer.

1. The acceleration is \(a(t)=v'(t)=(1.6t-0.08t^2)e^{-0.1t}\). 2. Critical points of \(a\) satisfy \(a'(t)=v''(t)=(0.008t^2-0.32t+1.6)e^{-0.1t}=0\). This gives \(t=20\pm10\sqrt{2}\). 3. Compare \(a(t)\) at these two critical points and at the endpoints \(t=0\) and \(t=60\). The maximum occurs at \(t=20-10\sqrt{2}\approx5.86\). 4. At that time, \(a(t)\approx3.69\,\text{m/s}^2\).

Approximately \(3.69\,\text{m/s}^2\)