Problem 32
Question
In Problems 29-34, sketch the graph of a continuous function fon \([0,6]\) that satisfies all the stated conditions. \(f(0)=3 ; f(2)=2 ; f(6)=0\); $$ \begin{aligned} &f^{\prime}(x)<0 \text { on }(0,2) \cup(2,6) ; f^{\prime}(2)=0 \\ &f^{\prime \prime}(x)<0 \text { on }(0,1) \cup(2,6) ; f^{\prime \prime}(x)>0 \text { on }(1,2) \end{aligned} $$
Step-by-Step Solution
Verified Answer
The graph is decreasing on \((0, 2)\) and \((2, 6)\) with a horizontal tangent at \(x = 2\).
1Step 1: Understand the Initial Values
Identify the specific points given by the initial condition: \(f(0) = 3\), \(f(2) = 2\), and \(f(6) = 0\). These will be critical points on the graph.
2Step 2: Analyze the Derivative Conditions
The condition \(f'(x) < 0\) means the function is decreasing on intervals \((0, 2)\) and \((2, 6)\). The condition \(f'(2) = 0\) indicates there is a horizontal tangent at \(x = 2\).
3Step 3: Analyze the Second Derivative Conditions
For \(f''(x) < 0\) on intervals \((0, 1)\) and \((2, 6)\), the graph is concave down. For \(f''(x) > 0\) on interval \((1, 2)\), the graph is concave up.
4Step 4: Sketch the Graph
Begin at point \((0, 3)\). Draw a decreasing and concave down segment from \(x = 0\) to \(x = 1\). Change the concavity to concave up at \(x = 1\) until \(x = 2\), maintaining the decreasing slope, and ending at \((2, 2)\) with a horizontal tangent. Continue to \(x = 6\) with a decreasing, concave down segment, ending at \((6, 0)\).
Key Concepts
Derivative conditionsSecond derivative analysisConcavityFunction behavior
Derivative conditions
Derivative conditions help us understand how a function behaves over certain intervals. In calculus, the first derivative of a function, denoted as \(f'(x)\), gives us information about the rate of change of the function, frequently seen as its slope. When \(f'(x) < 0\), it indicates that the function is decreasing over that interval. This means the function's values are getting smaller as the input \(x\) increases.
In the original problem, the derivative condition \(f'(x) < 0\) on intervals \((0, 2)\) and \((2, 6)\) tells us that the function is decreasing in these regions. Specifically, this means as we move left to right from 0 to 2, and then from 2 to 6, the graph of the function should be moving downwards. Additionally, the condition \(f'(2) = 0\) suggests there's a horizontal tangent at \(x = 2\). This generally implies a peak or a trough, or a point of inflection, depending on second derivative conditions, which we'll explore next.
In the original problem, the derivative condition \(f'(x) < 0\) on intervals \((0, 2)\) and \((2, 6)\) tells us that the function is decreasing in these regions. Specifically, this means as we move left to right from 0 to 2, and then from 2 to 6, the graph of the function should be moving downwards. Additionally, the condition \(f'(2) = 0\) suggests there's a horizontal tangent at \(x = 2\). This generally implies a peak or a trough, or a point of inflection, depending on second derivative conditions, which we'll explore next.
Second derivative analysis
Second derivative analysis provides insights into the concavity of the function's graph. The second derivative, \(f''(x)\), tells us how the rate of change of the slope itself is changing. In simpler terms, it helps us determine whether a curve is concave up (like a cup) or concave down (like an arch).
If \(f''(x) < 0\), the graph is concave down; this means the graph is curved like an upside-down bowl. Conversely, if \(f''(x) > 0\), the graph is concave up, indicating a cup-like curve. In the given exercise, it is stated that \(f''(x) < 0\) for intervals \((0, 1)\) and \((2, 6)\), and \(f''(x) > 0\) for \((1, 2)\). This implies the graph starts concave down from \(x = 0\) to \(x = 1\), changes to concave up from \(x = 1\) to \(x = 2\), and then returns to concave down from \(x = 2\) to \(x = 6\). These variations in concavity assist in accurately sketching the curve.
If \(f''(x) < 0\), the graph is concave down; this means the graph is curved like an upside-down bowl. Conversely, if \(f''(x) > 0\), the graph is concave up, indicating a cup-like curve. In the given exercise, it is stated that \(f''(x) < 0\) for intervals \((0, 1)\) and \((2, 6)\), and \(f''(x) > 0\) for \((1, 2)\). This implies the graph starts concave down from \(x = 0\) to \(x = 1\), changes to concave up from \(x = 1\) to \(x = 2\), and then returns to concave down from \(x = 2\) to \(x = 6\). These variations in concavity assist in accurately sketching the curve.
Concavity
Concavity is a crucial characteristic of a graph because it gives us important information about the shape of the graph over different intervals. Understanding concavity helps identify regions where the graph is curving upwards or downwards.
From the second derivative analysis, we know:
From the second derivative analysis, we know:
- The graph is concave down over \((0, 1)\) and \((2, 6)\).
- The graph is concave up over \((1, 2)\).
Function behavior
Understanding function behavior encapsulates the combined effect of derivative and concavity analyses. By considering given conditions like the values of the function at particular points, and derivative signs, we can predict and sketch how the function behaves from start to end.
In the given exercise:
In the given exercise:
- The function starts at a height of 3 at \(x = 0\), decreases to 2 at \(x = 2\), and then further decreases to 0 at \(x = 6\).
- In the interval \((0, 2)\), it is initially concave down until \(x = 1\), suggesting a typical upside-down bowl shape, and then concave up towards \(x = 2\).
- From \(x = 2\) onward till \(x = 6\), the function resumes its decreasing, concave down behavior until reaching its lowest point at \(x = 6\).
Other exercises in this chapter
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