Toolkit behavior, then absolute maximum and minimum

Let f(x) = -12$\frac{1}{2}$x + 3 with domain -4 ≤ x < 2. On the graph, the left end at (-4, 5) is a filled dot, so x = -4 is included. The right end at (2, 2) is a hollow dot, so x = 2 is excluded. The graph is a straight line segment with slope -12$\frac{1}{2}$ and no other pieces. Find the absolute maximum and the absolute minimum of f, if they exist. For each one, give the value and every input where it occurs.

1. Identify the toolkit shape. f(x) = -12$\frac{1}{2}$x + 3 is a linear function with slope -12$\frac{1}{2}$ and y-intercept 3.A line has no turning point, so its highest and lowest heights on a restricted domain can only occur at the ends of that domain.
2. Read the whole stated domain, -4 ≤ x < 2. The input x = -4 is included because it has a filled dot. The input x = 2 is excluded because it has a hollow dot. There are no arrows, so the graph stops at both ends.The domain decides which inputs are accepted. The edges of the viewing window do not set the domain. The stated inequality and the dots do.
3. Check for unbounded behavior. The domain is a bounded interval and the line is continuous on it. So the outputs neither rise nor fall without bound.Unbounded outputs would rule out an extremum. Here that does not happen, so we compare the heights at the ends.
4. Compute the end heights. f(-4) = -12$\frac{1}{2}$(-4) + 3 = 2 + 3 = 5. f(2) = -12$\frac{1}{2}$(2) + 3 = -1 + 3 = 2.These are the only candidates for extremes on a line segment.
5. The slope is negative, so f decreases from left to right. For every accepted x with -4 ≤ x < 2, we have 2 < f(x) ≤ 5.A decreasing function gives its greatest output at the leftmost input and smaller outputs further right.
6. Maximum: the height 5 is reached at the included input x = -4, and f(x) ≤ 5 for every domain input. No other input gives 5, because the line is strictly decreasing.An absolute maximum must be attained at an accepted input and be at least as large as every output.
7. Minimum: the height 2 occurs only at x = 2, which is excluded. Every accepted input gives an output strictly greater than 2. For any reached height f(x₀) > 2, an input slightly to the right of x₀ gives a smaller output.A height that is approached but never reached is not an absolute extremum. No reached output is the least, so there is no absolute minimum.

For f(x) = x, give behavior and extrema.

1. The output increases whenever the input increases.Output and input are the same number.
2. There are no local turns or flat comparisons and no absolute record heights.Every input has a larger-output neighbor on one side and smaller-output neighbor on the other; the domain extends without bound.

For f(x) = |x|, give its behavior and absolute extrema.

1. The outputs at −3, 0, and 2 are 3, 0, and 2.Absolute value is distance from zero.
2. Report decreasing before 0 and increasing after 0.The picture falls toward its vertex, then rises.
3. The absolute minimum is 0 at x = 0; there is no maximum.Distances cannot be negative, zero is reached, and arbitrarily large inputs have arbitrarily large distances.

Let f(x) = -(x - 2)²${)}^2$ + 5 with domain -1 < x ≤ 4. On the graph, the left end at (-1, -4) is a hollow dot, so x = -1 is excluded. The right end at (4, 1) is a filled dot, so x = 4 is included. The vertex is at (2, 5). Find the absolute maximum and the absolute minimum of f, if they exist, and state every input where each occurs.

1. Identify the toolkit shape. f(x) = -(x - 2)²${)}^2$ + 5 is the squaring toolkit function reflected over the x-axis, shifted right 2 and up 5. It is a downward-opening parabola with vertex (2, 5).The squaring toolkit accepts every real input. Only the stated domain restricts this problem, so we know where the vertex is and which way the parabola opens.
2. Read the whole stated domain: -1 < x ≤ 4. The domain has no arrows and is bounded on both sides. The left endpoint x = -1 is excluded (hollow dot). The right endpoint x = 4 is included (filled dot).Absolute extrema are judged only over the domain. The edges of the viewing window are not the domain endpoints; the stated inequality is.
3. Check for unbounded behavior. The domain is a bounded interval, and every output satisfies f(x) ≤ 5. Outputs do not rise without bound, and they do not fall without bound.Unbounded rising rules out an absolute maximum, and unbounded falling rules out an absolute minimum. Neither happens here, so we must check reached heights.
4. Check the excluded endpoint. Compute f(-1) = -(-3)²${)}^2$ + 5 = -9 + 5 = -4. Since x = -1 is not in the domain, the height -4 is approached as x moves toward -1 from the right but is never reached.A height that is approached but never attained cannot be an absolute extremum.
5. Find the greatest reached output. For every x, (x - 2)²${)}^2$ ≥ 0, so -(x - 2)²${)}^2$ + 5 ≤ 5. Equality holds only when x = 2, and x = 2 is in the domain. So f(2) = 5 ≥ f(x) for every domain input, and no other input gives 5.The absolute maximum must be attained at an accepted input and must be at least every other output. The vertex meets both conditions.
6. Look for the least reached output. The other candidate is the included endpoint: f(4) = -(2)²${)}^2$ + 5 = 1. But inputs near -1 give outputs below 1. For example, f(-0.9) = -8.41 + 5 = -3.41 and f(-0.99) = -8.9401 + 5 = -3.9401.Every candidate for the absolute minimum must be compared with all domain inputs, not just the endpoints and the vertex.
7. Conclude about the minimum. Every reached output is greater than -4. For any reached value y > -4, an input closer to -1 gives a smaller output. So no reached output is less than or equal to all others.The only possible floor, -4, is unattained. Without an attained least height, there is no absolute minimum.

Let f(x) = 14$\frac{1}{4}$(x + 1)²${)}^2$ + 2 with domain -4 < x ≤ 5. The graph is a wide upward-opening parabola with vertex (-1, 2). Near the vertex the curve looks almost flat: f(-2) = 2.25 and f(0) = 2.25. The left end at (-4, 4.25) is a hollow dot, so x = -4 is excluded. The right end at (5, 11) is a filled dot, so x = 5 is included. Find the absolute maximum and the absolute minimum of f, if they exist, and give every input where each one occurs.

1. Identify the toolkit shape: f is the squaring function x²$x^2$, shifted 1 left and 2 up, and vertically compressed by a factor of 14$\frac{1}{4}$. It opens upward with vertex (-1, 2).The form a(x - h)²${)}^2$ + k with a = 14$\frac{1}{4}$ > 0, h = -1, k = 2 gives an upward parabola whose vertex is (h, k).
2. Read the whole domain -4 < x ≤ 5. It is a bounded interval. The left end x = -4 is excluded (hollow dot). The right end x = 5 is included (filled dot). Nothing continues past either end.Only accepted inputs can produce an extreme value, so the endpoint dots must be checked first.
3. Check for unbounded behavior. The domain is bounded, so the outputs cannot rise or fall without bound. Both extrema may exist.An extremum fails to exist by unboundedness only when the outputs grow or drop without limit on the domain.
4. Find the lowest height. The vertex input x = -1 lies in the domain, and f(-1) = 14$\frac{1}{4}$(0)²${)}^2$ + 2 = 2. Solve f(x) = 2: 14$\frac{1}{4}$(x + 1)²${)}^2$ = 0, so x = -1 is the only solution. The nearby values f(-2) = 2.25 and f(0) = 2.25 are close to 2 but larger.The flat look near the vertex is misleading. The squared term is positive for every x ≠ -1, so the height 2 is reached at one input only.
5. Find the highest height. f increases as x moves away from -1. To the left, the farthest input is near -4, where f(-4) = 14$\frac{1}{4}$(9) + 2 = 4.25, but x = -4 is excluded. To the right, f(5) = 14$\frac{1}{4}$(36) + 2 = 9 + 2 = 11, and x = 5 is included.The hollow dot at height 4.25 is not attained, but it is lower than 11 anyway. The included right end gives the greatest output.
6. Compare the reached heights across the whole domain. Every output satisfies 2 ≤ f(x) ≤ 11. The value 2 occurs only at x = -1. The value 11 occurs only at x = 5, because the only other solution of (x + 1)²${)}^2$ = 36 is x = -7, which lies outside the domain.An absolute extremum is a reached output that is at least as large (or at least as small) as every other output, together with all inputs that produce it.

Read behavior and extrema of f(x) = 1x$\frac{1}{x}$.

1. The negative branch moves from f(−2) = −12$\frac{1}{2}$ to f(−1) = −1.These input moves stay on one side of the excluded zero.
2. The positive branch moves from f(1) = 1 to f(2) = 12$\frac{1}{2}$.The larger positive denominator gives a smaller positive fraction.
3. Report two decreasing intervals and no extrema.Each branch strictly falls, and outputs are unbounded above and below near the missing input.

Let f(x) = 6x$\frac{6}{x}$ with domain x ≥ 1. The left end at (1, 6) is a filled dot, so x = 1 is included. The domain continues to the right without end. A graphing window happens to stop at x = 12, but that is only where the picture is cut off. A table of values gives: f(1) = 6, f(2) = 3, f(3) = 2, f(6) = 1, f(12) = 12$\frac{1}{2}$. Find the absolute maximum and the absolute minimum of f, if they exist, and give every input where each occurs.

1. Identify the toolkit shape: f(x) = 6x$\frac{6}{x}$ is the reciprocal function 1x$\frac{1}{x}$ stretched vertically by 6. On x > 0 it is decreasing, and as x grows its outputs approach 0 from above.Knowing the toolkit behavior tells us how the graph acts beyond the plotted points. The reciprocal has the horizontal asymptote y = 0.
2. Read the whole stated domain: x ≥ 1. It has an included left endpoint at x = 1 and no right endpoint. The window edge at x = 12 is not a domain endpoint.Extrema are judged over every accepted input, not only the visible part of the graph.
3. Check for unbounded behavior. On x ≥ 1 the outputs never rise above 6 and never fall below 0, so they are bounded in both directions.An absolute maximum is ruled out automatically only when the outputs rise without bound. An absolute minimum is ruled out automatically only when they fall without bound.
4. Test the low level 0. Solving 6x$\frac{6}{x}$ = 0 has no solution, because a fraction with numerator 6 is never 0. For any input a, the larger input 2a gives f(2a) = 3a$\frac{3}{a}$, which is less than f(a). So no reached output is the least.A height that is approached but never attained is not an absolute minimum. Since every reached height has a lower one beyond it, there is no absolute minimum.
5. Find the greatest reached output. f is decreasing on the domain, so the largest output occurs at the smallest accepted input, x = 1. That output is f(1) = 6. Every other input x > 1 gives 6x$\frac{6}{x}$ < 6, so x = 1 is the only location.The value 6 is attained at the included endpoint, and f(1) ≥ f(x) for every x ≥ 1.

For f(x) = x$\sqrt{x}$, give the domain, the course's increasing interval, and extrema.

1. Accept inputs x ≥ 0, including 0.A real square cannot be negative and 0$\sqrt{0}$ = 0.
2. Read the rising graph and report increasing on (0, ∞).The course uses open behavior intervals, while the actual function also increases when comparing 0 with a larger input.
3. The absolute minimum is 0 at x = 0; there is no absolute maximum or local extremum.All outputs are nonnegative, 0 is reached, the curve continues upward without bound, and its endpoint has no domain neighborhood on both sides.

Read behavior and extrema of f(x) = ³x$\sqrt[3]{x}$.

1. f(−8) = −2, f(0) = 0, and f(27) = 3.(−2)³${)}^3$ = −8, 0³$0^3$ = 0, and 3³$3^3$ = 27.
2. Report increasing on (−∞, ∞), with no extrema.The curve keeps rising with no turn, level interval, or upper or lower output bound.

Let f(x) = -x - 2$\sqrt{x-2}$ + 6 with domain 2 < x ≤ 11. On the graph, the left end at (2, 6) is a hollow dot, so x = 2 is excluded. The right end at (11, 3) is a filled dot, so x = 11 is included. A table of values from the domain gives f(3) = 5, f(6) = 4 and f(11) = 3. Find the absolute maximum and the absolute minimum of f, if they exist. For each one that exists, give the value and every input where it occurs. If one does not exist, explain why.

1. Identify the toolkit shape. The function f(x) = -x - 2$\sqrt{x-2}$ + 6 is the square-root toolkit function x$\sqrt{x}$ shifted right 2, reflected across the horizontal axis, then shifted up 6.x$\sqrt{x}$ rises on x ≥ 0. The minus sign flips the graph vertically, so f falls steadily as x increases. A function that always falls has no turning point in the middle of its domain.
2. Read the whole stated domain: 2 < x ≤ 11. The left end x = 2 is excluded (hollow dot at (2, 6)). The right end x = 11 is included (filled dot at (11, 3)). There are no arrows, so the graph does not continue past either end.An extremum must be attained at an input that is in the domain. Included and excluded ends must be noted before any heights are compared.
3. Check for unbounded behavior. On 2 < x ≤ 11, the value x - 2$\sqrt{x-2}$ lies between 0 and 3. So f(x) lies between 6 - 3 = 3 and 6 - 0 = 6.The outputs are trapped between 3 and 6. They neither rise nor fall without bound, so unboundedness does not rule out either extremum.
4. Test the top level, 6. As x moves toward 2 from the right, f(x) approaches 6. Reaching 6 would need x - 2$\sqrt{x-2}$ = 0, which means x = 2, and x = 2 is excluded. For every x in the domain, x - 2$\sqrt{x-2}$ > 0, so f(x) < 6. Any accepted input x has a slightly smaller accepted input that gives a larger output.A height that is approached but never reached is not an absolute maximum. The hollow dot at (2, 6) marks an unattained level, and no reached output is the largest. So there is no absolute maximum.
5. Test the bottom level, 3. At x = 11, f(11) = -9$\sqrt{9}$ + 6 = -3 + 6 = 3, and x = 11 is in the domain. For every x in the domain, x - 2$\sqrt{x-2}$ ≤ 9$\sqrt{9}$ = 3, so f(x) ≥ 3. Equality needs x - 2$\sqrt{x-2}$ = 3, which happens only at x = 11.The value 3 is reached at an accepted input, and no domain input gives a lower output. That makes 3 the absolute minimum, and x = 11 is the only input that produces it.
6. Compare with the table: f(3) = 5, f(6) = 4, f(11) = 3. These outputs fall as x grows, all are at least 3, and all are below 6.Reached heights at sample inputs agree with the analysis: a falling shape with its lowest point at the included right end.

Let f(x) = -(x + 2)²${)}^2$ + 7 with domain -5 ≤ x ≤ 0. The graph is a downward-opening parabola with vertex (-2, 7). The left end at (-5, -2) is a filled dot, so x = -5 is included. The right end at (0, 3) is a filled dot, so x = 0 is included. Find the absolute maximum and the absolute minimum of f, and state where each occurs.

1. Identify the toolkit shape. f(x) = -(x + 2)²${)}^2$ + 7 is the squaring function x²$x^2$ reflected over the x-axis, shifted 2 left and 7 up. It is a downward-opening parabola with vertex (-2, 7).The squaring toolkit function accepts every real input. Here the stated domain -5 ≤ x ≤ 0 is the only restriction.
2. Read the whole domain: -5 ≤ x ≤ 0, with filled dots at both ends and no arrows.Filled dots mean x = -5 and x = 0 are accepted inputs. The graph stops there, so the outputs cannot rise or fall without bound.
3. Check for heights that are approached but never reached. Both endpoints are included and the curve is unbroken between them, so there are no hollow dots or unattained levels.Only heights reached at accepted inputs can be absolute extrema.
4. Find the greatest output. Since (x + 2)²${)}^2$ ≥ 0, f(x) = 7 - (x + 2)²${)}^2$ ≤ 7 for every x. The value 7 is reached when x + 2 = 0, which gives x = -2, and -5 ≤ -2 ≤ 0.The vertex is the highest point of a downward parabola, and here it lies inside the domain.
5. Find the least output. Evaluate both endpoints: f(-5) = -(-3)²${)}^2$ + 7 = -9 + 7 = -2 and f(0) = -(2)²${)}^2$ + 7 = -4 + 7 = 3. Compare them: -2 < 3.A downward parabola decreases as you move away from the vertex. The lowest point on a closed interval is therefore at the endpoint farther from the vertex. That endpoint is x = -5, which is 3 units from x = -2, while x = 0 is only 2 units away.
6. Confirm that -2 is the least reached output and find every input that gives it. On the domain, -3 ≤ x + 2 ≤ 2, so (x + 2)²${)}^2$ ≤ 9 and f(x) ≥ -2. Equality needs x + 2 = ±3, so x = -5 or x = 1. Only x = -5 is in the domain.The location must list every accepted input that produces the extreme value.

For f(x) = x on domain [0, 2), find the absolute maximum and minimum.

1. f(0) = 0 and every accepted output is at least 0.Every accepted input is nonnegative, and the output equals the input.
2. No accepted input produces output 2.The right endpoint is excluded.
3. For any accepted input b, the input b + 22$\frac{b+2}{2}$ is still below 2 and is greater than b.It is halfway from b toward 2, so it gives a larger accepted output.
4. There is no absolute maximum; the absolute minimum is 0 at 0.Every candidate high output is beaten, while the lower bound 0 is reached.

The given full curve f(x) = x³$x^3$ − 12x rises to (−2, 16), falls to (2, −16), then rises again. Its left end continues downward without bound and its right end upward without bound. Explain its local and absolute extrema.

1. There is local maximum 16 at −2 and local minimum −16 at 2.The stated direction changes make the nearby two-sided comparisons hold.
2. There is no absolute maximum or minimum.The given end behavior beats every upper or lower candidate bound.
3. State that local describes a neighborhood while absolute describes the whole domain.The turning values win their nearby comparisons but lose the whole-domain comparison.

Let the gas-price function's domain be exactly the eight recorded years in the picture. Find its absolute maximum and minimum and describe the adjacent-year peak and dip honestly.

1. Scan every output column; 3.68 is greatest and 2.31 is least.All domain outputs appear in this finite table.
2. The column under 2012 gives 3.68 and the column under 2005 gives 2.31.These are the input locations of the record outputs.
3. The 2008 output 3.30 exceeds its adjacent recorded outputs 2.84 and 2.41; the 2009 output 2.41 is below adjacent 3.30 and 2.84.Those column comparisons describe an observed sampled peak and dip, rather than continuous behavior at unrecorded times.