Free math worksheets

- 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: \(f(-1)=-2\).

- On a closed interval, the candidates are the endpoints plus every interior critical number. - Differentiate to find the interior critical number. - Do not identify the absolute maximum until you have compared the function value at every candidate. - After the candidate comparison, use the sign of \(P'\) to state the monotonicity intervals.

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 \(512\) at \(x=8\), corresponding to \(8000\) revolutions per minute. The absolute minimum value is \(0\), 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 \(512\) at \(8000\) revolutions per minute. Increasing on \([0,8]\); decreasing on \([8,12]\).

- 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. - The polyline has corners at the interior turning vertices. - For the candidates test, include both endpoints and all four 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. Local maxima occur at \((3,5)\) and \((9,4)\). Local minima occur at \((6,0.5)\) 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. For the candidates test, compare \(h(0)=2\), \(h(3)=5\), \(h(6)=0.5\), \(h(9)=4\), \(h(12)=1\), and \(h(13)=1.5\). The smallest candidate value is \(0.5\), attained at \(t=6\).

a) Increasing on \([0,3]\), \([6,9]\), and \([12,13]\); decreasing on \([3,6]\) and \([9,12]\). b) Local maxima: \((3,5)\), \((9,4)\); 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 \(x=2\).

- 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 \(x = 3\)

- The denominator of \(f'\) is always positive, so only the factored numerator controls its sign. - Use the critical numbers to divide the real line into maximal derivative-sign intervals. - After finding the monotonicity, compare the critical-point values with the limiting value as \(x\to\pm\infty\).

1. Since \(x^2+3>0\) for every real \(x\), the function is continuous on \(\mathbb{R}\). 2. Differentiate: \(f'(x)=\frac{6(x-3)(x+1)}{(x^2+3)^2}\). The denominator is positive, so the sign of \(f'\) is the sign of \((x-3)(x+1)\). 3. Thus \(f'>0\) on \(( -\infty,-1)\), \(f'<0\) on \((-1,3)\), and \(f'>0\) on \((3,\infty)\). Therefore, \(f\) is increasing on \(( -\infty,-1)\) and \((3,\infty)\), and decreasing on \((-1,3)\). 4. The function values at the critical numbers are \(f(-1)=4\) and \(f(3)=0\). Also, \(\lim_{x\to\pm\infty}f(x)=1\). 5. By continuity and the monotonicity from step 3, \(0\) is the absolute minimum, \(4\) is the absolute maximum, and every value between them occurs. The range is \([0,4]\).

a) Increasing on \(( -\infty,-1)\) and \((3,\infty)\); decreasing on \((-1,3)\). b) Range: \([0,4]\).

- Use the sign of the first derivative to locate the absolute maximum. - Find the limit at both ends of the real line. - Decide whether the limiting value is ever attained.

1. Differentiate: \(h'(x)=\frac{2x(x^2+1)-2x(x^2+4)}{(x^2+1)^2}=\frac{-6x}{(x^2+1)^2}\). 2. The derivative is positive for \(x<0\) and negative for \(x>0\). Thus, \(h\) increases on \((-\infty, 0)\) and decreases on \((0, \infty)\). 3. The function has an absolute maximum at \(x=0\), where \(h(0)=4\). 4. Since \(\lim_{x\to\pm\infty}h(x)=1\), the value \(1\) is a lower bound. It is not attained because \(\frac{x^2+4}{x^2+1}>1\) for every real \(x\). 5. Therefore, the range is \((1, 4]\).

\((1, 4]\)

- 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 \(3\,\text{m/s}^2\).

- First inventory the candidates: endpoints and interior turning points. - Read one function value for every candidate before comparing them. - The absolute-extremum conclusion must come from the finite candidate table, not from scanning the whole graph.

1. The endpoints are \(x=-3,3\). The interior turning points are approximately \(x=-2,0,2\). Thus the candidates are \(-3,-2,0,2,3\). 2. From the graph, the candidate values are approximately \(f(-3)=1.9\), \(f(-2)=-0.6\), \(f(0)=1.0\), \(f(2)=-0.6\), and \(f(3)=1.9\). 3. The largest candidate value is about \(1.9\), so the absolute maxima occur at \(x=-3\) and \(x=3\). The smallest candidate value is about \(-0.6\), so the absolute minima occur at \(x=-2\) and \(x=2\). 4. The graph decreases on \([-3,-2]\) and \([0,2]\).

a) Candidates: approximately \((-3,1.9),(-2,-0.6),(0,1.0),(2,-0.6),(3,1.9)\). b) Absolute maxima: about \(1.9\) at \(x=-3,3\); absolute minima: about \(-0.6\) 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 only to locate interior candidates. - Do not forget the endpoints when building the complete candidate set. - The function-value table, not the derivative graph, determines which candidate is absolute.

1. The graph of \(f'\) has zeros at \(x=-2,1,3\), so these are the interior critical numbers. 2. The complete candidate set is \(x=-4,-2,1,3,4\). 3. The largest listed function value is \(5\) at \(x=-2\), and the smallest is \(-1\) at \(x=1\).

Interior critical numbers: \(x=-2,1,3\). Absolute maximum: \(5\) 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 \(3.69\,\text{m/s}^2\)

- Compare \(f_k(-x)\) with \(f_k(x)\). - Find and classify the critical points using the first derivative. - Compare numerator and denominator degrees for end behavior. - Use the absolute extrema to determine the range.

1. Since \(f_k(-x)=-f_k(x)\), every graph is symmetric about the origin. 2. The first derivative is \(f_k'(x)=\frac{k(4-x^2)}{(x^2+4)^2}\), so the critical points are \(x=-2\) and \(x=2\). 3. Their function values are \(f_k(-2)=-\frac{k}{4}\) and \(f_k(2)=\frac{k}{4}\). If \(k>0\), \(\left(-2, -\frac{k}{4}\right)\) is a local minimum and \(\left(2, \frac{k}{4}\right)\) is a local maximum. If \(k<0\), the classifications reverse. 4. Since the denominator has greater degree than the numerator, \(\lim_{x\to\pm\infty}f_k(x)=0\). 5. The function is continuous on \(\mathbb{R}\), and the two critical values are its absolute minimum and maximum. Therefore, the range is \(\left[-\frac{|k|}{4}, \frac{|k|}{4}\right]\).

1) Symmetric about the origin 2) Critical points: \(\left(-2, -\frac{k}{4}\right)\) and \(\left(2, \frac{k}{4}\right)\); their minimum/maximum roles depend on the sign of \(k\). 3) \(\lim_{x\to\pm\infty}f_k(x)=0\) 4) Range: \(\left[-\frac{|k|}{4}, \frac{|k|}{4}\right]\)