Tasks:
In Section 7.3.2, we see a new definition: concave up or concave down, in Definition 7.19.
A function is concave down on an interval [a,b] if the graph of the function lies below the tangent line to the graph at any point x in [a,b].
A function is concave up on an interval [a,b] if the graph of the function lies above the tangent line to the graph at any point x in [a,b].
We also have a special name for points where a function changes from concave up to down, or vice-versa. These are the inflection points of the function.

❗We will usually rely on a table of signs of the second derivative to tell us the concavity of a function.
This video contains the same details, as well as a visual explanation of concavity.

Question:

The graph of a function y=h(x) on the domain [0,4] is shown below. On which interval(s) is h(x) concave down?

Question

Tasks:
In Section 7.3.2, we see a new definition: concave up or concave down, in Definition 7.19.
A function is concave down on an interval [a,b] if the graph of the function lies below the tangent line to the graph at any point x in [a,b].
A function is concave up on an interval [a,b] if the graph of the function lies above the tangent line to the graph at any point x in [a,b].
We also have a special name for points where a function changes from concave up to down, or vice-versa. These are the inflection points of the function.

❗We will usually rely on a table of signs of the second derivative to tell us the concavity of a function.
This video contains the same details, as well as a visual explanation of concavity.

Question:

The graph of a function y=h(x) on the domain [0,4] is shown below. On which interval(s) is h(x) concave down?

BlackTom AI · Accepted Answer

To determine where h(x) is concave down on the domain [0,4], you would normally examine the second derivative or, graphically, check where the graph lies below the tangent lines at points x in the interval. Concavity down corresponds to h''(x) < 0, which means the slope of h'(x) is decreasing as x increases. Key approach you would use for each option: - For any subinterval, if the graph lies below the tangent line at each point x in that subinterval, then h is concave down there. Equivalently, the second derivative should be negative across that subinterval. - If you have a sign table or information about h''(x) from the material (where h''(x) < 0 indicates concave down and h''(x) > 0 indicates concave up), you would identify the portions of [0,4] where h''(x) is negative and then intersect those with the proposed interval(s). - If the graph has inflection points (where concavity changes), those points split the domain into regions of consistent concavity. In the absence of exact visual data or a provided second-derivative sign chart, you cannot definitively label a subinterval as concave down. Now, evaluating each option in principle: Option A: 1 < x < 2 and 3 < x < 4 - This option asserts concavity down on two separate subintervals. If the graph’s concavity changes within [0,4], these would be plausible regions where h''(x) < 0. To confirm, you would check on 1 < x < 2 and on 3 < x < 4 individually whether the graph lies below the tangent lines at representative points (or whether h''(x) is negative throughout these ranges). Without the graph, you cannot be sure these are indeed negative-curvature regions. Option B: 0 < x < 2 - This claims concavity down across a longer span from 0 to 2. If the graph is consistently bending downward on that entire stretch, then this would be correct. However, if there is any subsegment in 0 < x < 2 where the graph bends upward (concave up), then this option would be incorrect. This hinges on whether a sign change in concavity occurs inside (0,2). Option C: 2 < x < 4 - This suggests concavity down on the second half of the interval. Similar logic applies: confirm whether h''(x) < 0 for all x in (2,4). If there is an inflection point between 2 and 4 or a portion where the curvature is upward, this option would be invalid. Without the image or a sign chart, a determination cannot be made. Option D: 0 < x < 1 and 2 < x < 3 - This presents two disjoint subintervals. It would be correct if h''(x) < 0 on both (0,1) and (2,3) while possibly being nonnegative elsewhere. Again, the key is consistency of negative second derivative across those subintervals, which you would verify from the graph by checking tangents or from the second-derivative sign information. Important note: the provided question refers to a specific graph of h(x) on [0,4], but the graph image is not available here. Without seeing how the graph lies relative to its tangents (or without the sign information for h''(x) over subintervals), we cannot definitively identify which interval(s) are concave down. If you have access to the graph or the second-derivative sign table, apply the steps above to each candidate interval, testing whether the curvature is downward throughout each subinterval. If a subinterval contains any point where the graph lies above its tangent line or where h''(x) > 0, that subinterval would not be concave down. In summary, to pick among the given options, you would: - determine where h''(x) < 0 on (0,4) from the graph or from h'' signs, - then see which of the listed subintervals are completely contained within those negative-sign regions, - and avoid selecting any interval that spans a region where concavity changes or where concavity is upward.

MAT133 F24-W25 W25 Week 6 Preclass Guided Reading

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