Problem 9 - Calculus Concepts

Question

Sketch a possible graph of \(t\) with input \(x\), given that \(t(3)=7\) \(t(4.4)=t(8)=0\) \(t^{\prime}(6.2)=0\) the graph of \(t\) has no concavity changes.

Step-by-Step Solution

Answer

Plot points, ensure concave down shape throughout, starting at (3,7), 0 at (4.4) and (8), flat at (6.2).

Step 1: Understand the Given Points

The function moves through specific points, so we start by plotting these. First, plot the point \((3, 7)\) since \(t(3) = 7\). This means that at \(x = 3\), the value of the function \(t\) is 7.

Step 2: Identify Zeros of the Function

Now, locate and plot the points where \(t(x) = 0\). These are \((4.4, 0)\) and \((8, 0)\). The function will cross the x-axis at these points.

Step 3: Determine the Slope at Critical Points

Since \(t'(6.2) = 0\), it means that the derivative of \(t\) at \(x = 6.2\) equals 0, indicating a horizontal tangent at this point. Mark \(x = 6.2\) on the x-axis where the graph has a flat tangent.

Step 4: Consider Continuity and General Shape

Since there are no concavity changes, \(t\) is either entirely concave up or concave down. Choose one consistent shape based on points and tangent behavior. Here, let's choose concave down for simplicity: rising steeply at \(x = 3\), reaching zero at \(x = 4.4\), creating a local maximum at \(x = 6.2\), and descending through \(x = 8\).

Step 5: Sketch the Graph

Sketch a graph using the plotted points and the identified curve type. Start at \((3, 7)\), dip to \((4.4, 0)\), rise to a peak at \(x = 6.2\) (the height consistent with concavity), and finally descend to cross the x-axis again at \((8, 0)\). Ensure the entire graph shows no change in concavity.

Key Concepts

Understanding Derivatives

In graph sketching, derivatives play a crucial role as they tell us about a function's rate of change at any given point. The derivative of a function, often represented as \( t'(x) \), provides a snapshot of how the function behaves locally:

• A positive derivative indicates the function is increasing.
• A negative derivative shows the function is decreasing.
• A zero derivative reveals a horizontal tangent, suggesting a potential local maximum or minimum.

In the given exercise, it is stated that \( t'(6.2) = 0 \). This tells us that at \( x = 6.2 \), the function has a horizontal tangent line. In our graph, this translates to either a peak (local maximum) or a trough (local minimum) or sometimes just a flat inflection point. However, other information is crucial to distinguish between these possibilities.

Exploring Curve Concavity

Concavity refers to the direction in which a curve bends. In function analysis, concavity gives insights into how steepness changes. If a function's second derivative \( t''(x) \) is:

• Positive, the function is concave up, resembling an upright bowl shape.
• Negative, the function is concave down, similar to an upside-down bowl.

In this exercise, we're informed there are no concavity changes throughout the graph of \( t \). This means that from start to end, the graph consistently curves in one way. We chose a concave-down direction for clarity, leading to a shape starting steep, flattening out at a point like \( x = 6.2 \), and descending once more. This constant bending indicates no shifts in the steepness direction, simplifying our graph’s shape. Such information is vital in accurately depicting the graph's long-run behavior.

Locating Critical Points

Critical points are where a function's derivative is zero or undefined, marking potential changes in the function's direction. These points are key in detailing the function's landscape. For our function \( t(x) \):

• We have a critical point at \( x = 6.2 \) because \( t'(6.2) = 0 \).
• At this point, based on the absence of concavity changes, it likely indicates a local maximum, where the function shifts from increasing to decreasing without changing the curvature.

Additionally, we pinpoint the function's zeros where \( t(x) = 0 \) at \( x = 4.4 \) and \( x = 8 \). These intersections with the x-axis represent where the function crosses, aiding in sketching the overall trajectory. By integrating these critical points with the information on derivatives and concavity, you build a thorough understanding of constructing an accurate and analytically sound graph.