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

- Find the critical numbers in the closed interval. - Use the sign of the first derivative to determine where the model increases or decreases. - Compare the function values at interior critical numbers and endpoints.

1. Differentiate: \(P'(x)=-6x^2+48x=-6x(x-8)\). 2. The candidates on \([0,12]\) are the endpoints \(x=0\), \(x=12\), and the interior critical number \(x=8\). 3. The derivative is positive on \((0,8)\) and negative on \((8,12)\). Therefore, \(P\) is strictly increasing on \([0,8]\) and strictly decreasing on \([8,12]\). 4. Compare the candidate values: \(P(0)=0\), \(P(8)=512\), and \(P(12)=0\). The absolute maximum occurs at \(x=8\), corresponding to \(8000\) revolutions per minute.

\(P\) is strictly increasing on \([0,8]\) and strictly decreasing on \([8,12]\). The maximum power occurs at \(8000\) revolutions per minute.

- Turning points separate increasing and decreasing intervals. - Read both coordinates of each peak and valley. - Include endpoints when finding the absolute minimum.

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 \((12, 1)\), where each ordered pair is \((\text{time in hours}, \text{level in meters})\). 3. Comparing the local extrema and endpoints, 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) Local maxima: \((3, 5)\) and \((9, 4)\); local minima: \((6, 0.5)\) and \((12, 1)\) c) The lowest water level 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.

- Compare the heights of all marked candidate points, not just the turning points. - Absolute extrema compare values across the entire displayed interval. - Relate the interior turning points to the derivative condition used in the candidates test.

1. The marked candidate values are \(f(-5)=4\), \(f(-3)=-2\), \(f(0)=3\), \(f(2)=-2\), and \(f(3)=1\). 2. The absolute maximum is \(4\) at \(x=-5\). 3. The absolute minimum is \(-2\), attained at \(x=-3\) and \(x=2\). 4. The candidates test checks endpoints and interior critical numbers; on this smooth graph, the marked turning points are the interior critical numbers.

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 \(x = 3\)

- Check whether the function is continuous on all real numbers. - Find and classify the critical points. - Compare the critical values with the end behavior. - Use continuity to determine whether all intermediate values occur.

1. Since \(x^2+3>0\) for every real \(x\), the function is continuous on \(\mathbb{R}\). 2. Differentiate: \(f'(x)=\frac{6x^2-12x-18}{(x^2+3)^2}=\frac{6(x-3)(x+1)}{(x^2+3)^2}\). 3. The critical points are \(x=-1\) and \(x=3\). The derivative changes from positive to negative at \(x=-1\), so \(f(-1)=4\) is a local maximum. It changes from negative to positive at \(x=3\), so \(f(3)=0\) is a local minimum. 4. Also, \(\lim_{x\to\pm\infty}f(x)=1\), which lies between \(0\) and \(4\). 5. Because the function is continuous on all real numbers, the minimum \(0\) and maximum \(4\) are absolute, and every value between them occurs. Therefore, the range is \([0, 4]\).

\([0, 4]\)

- 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 \((1, 4]\).

\((1, 4]\)

- Read each turning point using the grid. - An absolute extremum is the highest or lowest value on the entire interval. - A local extremum only compares values in a nearby neighborhood. - Include the endpoint values when determining absolute extrema.

1. The graph has local minima at \((-3, -2)\) and \((1, -2)\), and a local maximum at \((-1, 2)\). 2. On \([-4, 2]\), the smallest function value is \(-2\), so both \((-3, -2)\) and \((1, -2)\) are absolute minima as well as local minima. 3. The point \((-1, 2)\) is a local maximum, but it is not the absolute maximum because both endpoint values are greater than \(2\). 4. The absolute maxima on the interval occur at \(x=-4\) and \(x=2\).

a) Local minima: \((-3, -2)\) and \((1, -2)\); local maximum: \((-1, 2)\) b) The two local minima are also absolute minima. The local maximum at \((-1, 2)\) is not absolute. The absolute maxima occur at \(x=-4\) and \(x=2\).

- Convert a value of \(t\) to a clock time by counting hours after 6:00 a.m. - Find the critical numbers in the closed interval. - Compare the temperature at the critical number and the endpoints.

1. The graph rises on \([0, 8]\) and falls on \([8, 12]\). Thus, 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. 2. Differentiate: \(T'(t)=-0.15t^2+1.2t=-0.15t(t-8)\). The critical numbers in \([0, 12]\) are \(t=0\) and \(t=8\). 3. The derivative is positive on \((0, 8)\) and negative on \((8, 12)\), confirming the monotonic intervals. 4. Compare candidates: \(T(0)=5\), \(T(8)=17.8\), and \(T(12)=5\). Therefore, the absolute maximum occurs at \(t=8\), which is 2:00 p.m.

a) Increasing on \([0, 8]\), from 6:00 a.m. to 2:00 p.m.; decreasing on \([8, 12]\), from 2:00 p.m. to 6:00 p.m. b) \(T'(t)=-0.15t(t-8)\). The maximum temperature occurs at 2:00 p.m.

- Include endpoints when finding absolute extrema on a closed interval. - Read turning-point coordinates and endpoint heights from the graph. - Report graph-based numerical values as estimates.

1. The graph has a local maximum near \((0, 1.0)\) and local minima near \((-2, -0.6)\) and \((2, -0.6)\). 2. The graph decreases from \(x=-3\) to \(x=-2\), and from \(x=0\) to \(x=2\). Thus, it is strictly decreasing on \([-3, -2]\) and \([0, 2]\). 3. The endpoint values are approximately \(f(-3)=f(3)=1.9\), which are the absolute maxima. The absolute minimum value is approximately \(-0.6\), attained at \(x=-2\) and \(x=2\).

a) Local maximum: approximately \((0, 1.0)\); local minima: approximately \((-2, -0.6)\) and \((2, -0.6)\) b) \([-3, -2]\) and \([0, 2]\) c) Absolute maximum: approximately \(1.9\) at \(x=-3\) and \(x=3\); absolute minimum: approximately \(-0.6\) at \(x=-2\) and \(x=2\)

- 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.

- Identify the highest and lowest points on the entire displayed interval. - Absolute extrema compare all points in the interval. - Local extrema compare only nearby points. - Endpoints can be absolute extrema.

1. The endpoint values are equal and are the highest values on the interval, so \(x=-3\) and \(x=3\) are absolute maxima. 2. At \(x=-2\) and \(x=2\), the function value is \(0\), the lowest value on the interval. These are absolute minima and also local minima. 3. At \(x=0\), the graph has a local maximum with value \(4\), but this is not an absolute maximum because the endpoint values are larger.

Absolute maxima: \(x = -3\) and \(x = 3\) Absolute and local minima: \(x = -2\) and \(x = 2\) Local maximum: \(x = 0\)

- 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) The student 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 \(x=\pi\).

- 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: \(\left[-\frac{|k|}{4}, \frac{|k|}{4}\right]\)