Problem 19 - Calculus Early Transcendentals

Question

Designing a function Sketch a graph of a function \(f\) continuous on \([0,4]\) satisfying the given properties. \(f^{\prime}(x)=0\) for \(x=1\) and \(2 ; f\) has an absolute maximum at \(x=4 ; f\) has an absolute minimum at \(x=0 ;\) and \(f\) has a local minimum at \(x=2\).

Step-by-Step Solution

Answer

Question: Sketch a graph of a continuous function f on the closed interval from 0 to 4 with the following properties: 1. \(f'(x)=0\) at \(x=1\) and \(x=2\). 2. \(f\) has an absolute maximum at \(x=4\). 3. \(f\) has an absolute minimum at \(x=0\). 4. \(f\) has a local minimum at \(x=2\). Answer: Follow the step-by-step solution to sketch a graph satisfying the given conditions. Ensure the graph starts at an absolute minimum at \(x=0\) and gradually increases to reach an absolute maximum at \(x=4\), while having horizontal tangents at \(x=1\) and \(x=2\) and a local minimum at \(x=2\).

Step 1: Sketch a graph with the given critical points (x = 0, 1, 2, and 4)

Plot the points \((0, f(0))\), \((1, f(1))\), \((2, f(2))\), and \((4, f(4))\) on the graph. These points represent the critical points where the function has an absolute maximum, absolute minimum, and local minimum.

Step 2: Connect the critical points taking into account \(f'(x)=0\) at \(x=1\) and \(x=2\)

Since the function's derivative is 0 at \(x=1\) and \(x=2\), the curve must have horizontal tangents at these positions, i.e., they should be either local maxima or minima. From our given information, we know that a local minimum occurs at \(x=2\). So, connect the points in such a way that we have horizontal tangents at \(x=1\) and \(x=2\) while having a local minimum at \(x=2\).

Step 3: Include the absolute maximum at \(x=4\) and absolute minimum at \(x=0\)

According to the given properties, the function should have an absolute maximum at \(x=4\) and an absolute minimum at \(x=0\). So, the curve should begin at its lowest point and increase gradually, reaching its highest point at \(x=4\). Make sure the curve goes from the absolute minimum at \(x=0\) to absolute maximum at \(x=4\) while maintaining the local minimum at \(x=2\). The sketched graph should look like this: 1. Starts at the absolute minimum at \(x=0\). 2. The curve has a horizontal tangent at \(x=1\) (possibly a local maximum). 3. The curve has a horizontal tangent and local minimum at \(x=2\). 4. Ends at the absolute maximum at \(x=4\).

Key Concepts

Continuous Functions

A continuous function is one that smoothly connects all its points without any breaks, holes, or jumps. In simpler terms, imagine drawing the function's graph without lifting your pencil from the paper. A continuous function is essential in calculus because it reflects a type of function that behaves predictably within an interval, here \([0, 4]\).

• There are no interruptions between any two points on the curve.
• The y-values change smoothly as x-values change.

In the given exercise, the function \(f\) is continuous on the interval \([0,4]\). This means that as you move along the graph from \(x = 0\) to \(x = 4\), the function should not have any sharp edges or undefined points. Continuous functions ensure that calculating maximum and minimum values can be done without encountering unexpected results.

Critical Points

Critical points are the values of \(x\) where a function’s derivative is zero or undefined. These points are important because they can indicate where a function switches direction. In the context of the problem, critical points were found at \(x = 1\) and \(x = 2\) because \(f'(x) = 0\).

• Critical points are often areas where functions have peaks or valleys.
• They are potential locations for local maxima and minima.

To determine the nature of these critical points, examine the behavior of \(f(x)\) around them. For instance, the problem states there's a local minimum at \(x = 2\). This affects how we sketch the section of the graph near \(x = 2\), ensuring it dips into a local minimum. Knowing where these critical points lay helps in sketching a curve that accurately represents the function’s behavior.

Absolute Maximum

The absolute maximum of a function on a given interval is its highest value within that specified range. It's the tallest point on the graph restricted to the considered domain. In the exercise, the absolute maximum is at \(x = 4\).

• The absolute maximum is a pivotal point on the graph indicating where the function reaches its greatest output.
• No other point within the interval \([0,4]\) will have a greater y-value than at \(x = 4\).

In terms of sketching the function, this means that \(f(x)\) should rise from the lowest value at \(x = 0\) and reach its peak at \(x = 4\). Identifying the absolute maximum ensures the graph correctly portrays the function’s progression from low to high values within the interval.

Absolute Minimum

Conversely to an absolute maximum, the absolute minimum is the smallest output or y-value of the function within its domain. For the given problem, the absolute minimum occurs at \(x = 0\).

• The absolute minimum reflects the lowest point the function ever reaches on the interval \([0,4]\).
• No other point within this interval will yield a lower y-value than at \(x = 0\).

This characteristic is crucial because the function should begin at \(x = 0\) at its lowest point and then increase to reach the peak at the absolute maximum \(x = 4\). Identifying and understanding where the absolute minimum occurs aids in accurately sketching the initial slope of the function to convey the gradual increase to the maximum, while also respecting the local minima and other critical points highlighted in the exercise.