Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of a function \(f\) that has the following properties: (a) \(f\) is everywhere continuous; (b) \(f(-4)=-3, f(0)=0, f(3)=2\); (c) \(f^{\prime}(-4)=0, f^{\prime}(3)=0, f^{\prime}(x)>0\) for \(x<-4, f^{\prime}(x)>0\) for \(-43\) (d) \(f^{\prime \prime}(-4)=0, f^{\prime \prime}(0)=0, f^{\prime \prime}(x)<0\) for \(x<-4, f^{\prime \prime}(x)>0\) for \(-40\)

Step-by-Step Solution

Answer

Graph is a smooth curve passing through (-4, -3), (0, 0), and (3, 2), increasing and concave changes at specified intervals.

Step 1: Understand the Problem

We need to sketch a graph that fulfills four conditions related to continuity, specific values, and slopes/concavities at given points.

Step 2: Define Continuity

Since the function is everywhere continuous, it should have no jumps or breaks for any value of \(x\). Thus, our sketch can be a smooth curve.

Step 3: Plot Key Points

The function passes through three points: \((-4, -3)\), \((0, 0)\), and \((3, 2)\). These should be placed on the graph accurately.

Step 4: Calculate Derivative Information

From the derivative conditions, \(f'(x)>0\) implies the function is increasing, while \(f'(x)<0\) implies it is decreasing. Therefore, sketch the graph based on the following: increasing for \((-fty, -4), (-4, 3)\) and decreasing for \((3, fty)\). At \(-4\) and \(3\), the derivative is zero, suggesting a potential horizontal tangent or peak/trough.

Step 5: Incorporate Second Derivative Information

The second derivative, \(f''(x)\), indicates concavity: \(f''(x) < 0\) implies concave down, while \(f''(x) > 0\) implies concave up. Apply concave down for \((-fty,-4)\) and \((0, fty)\), and concave up for \((-4,0)\). At \(-4\) and \(0\), \(f''(x) = 0\), indicating possible inflection points.

Step 6: Sketch the Graph

Start at \((-4, -3)\) with a horizontal tangent (\(f'(-4)=0\)), increasing and concave down. At \(-4\) to \(0\), transition to concave up, passing through \((0, 0)\) with another tangent slope (\(f'(0)=0\)). From \(0\) to \(3, f(x)\) is still growing and changes concavity again, and at \(3\), there's a horizontal tangent. After \(3\), the curve decreases and remains concave down.

Key Concepts

Derivatives

Derivatives provide key information about the behavior and direction of a function. They tell us where a function is increasing or decreasing. For a given function \( f(x) \), the first derivative, \( f'(x) \), reveals the rate of change or slope of the tangent.

If \( f'(x) > 0 \), the function is increasing; if \( f'(x) < 0 \), it's decreasing.

In this exercise, we have derivative conditions:

• \( f'(-4)=0 \) and \( f'(3)=0 \): these points hint at potential horizontal tangents or peaks and troughs, similar to tops of hills or bottoms of valleys.
• For \( x < -4 \): \( f'(x) > 0 \), indicating the function increases
• From between \( -4 < x < 3 \): \( f'(x) > 0 \), the function continuously rises
• For \( x > 3 \): \( f'(x) < 0 \), implying the function decreases

Understanding derivatives is crucial for sketching the function's general direction and identifying where changes in its growth occur.

Concavity

Concavity describes the "bending" nature of a function. It helps determine how the slope changes as you move along the curve. The second derivative, \( f''(x) \), informs us about concavity:

• If \( f''(x) > 0 \), the function is concave up, like a cup, and the curve shapes upwards.
• If \( f''(x) < 0 \), it is concave down, resembles an upside-down cup, and the curve shapes downwards.

In our problem:

• \( f''(x) < 0 \) for \( x < -4 \), indicating a concave down behavior at the beginning
• \( -4 < x < 0 \) shows \( f''(x) > 0 \), creating a concave up section
• Lastly, \( f''(x) < 0 \) for \( x > 0 \), bringing back the concave down bending.

The changes in concavity at key points signal potential inflection points where the curvature changes direction.

Critical Points

Critical points involve any place where the first derivative is zero or undefined, providing potential shifts in the direction of a function's graph. In a continuous function like the one in this exercise, critical points often correspond to turning points, peaks, or troughs. Here are the critical features:

• At \( x = -4 \) and \( x = 3 \), \( f'(x) = 0 \), marking spots for horizontal tangents.
• These points are crucial as they signal transitions in the graph's rise and fall behavior

In the given function, the critical points coincide with its conditions for increasing and then decreasing. The knowledge of critical points supports us in mapping out where the curve peaks or dips. Hence, they become essential in graph sketching.

Graph Sketching

Graph sketching is visually representing the function's behavior based on its derivatives and other properties. This process combines all the insights gained from derivatives, concavity, and critical points.

• Start by plotting key points: \((-4, -3)\), \((0, 0)\), and \((3, 2)\), mapping where the curve should pass through.
• Apply the increase or decrease information from \( f'(x) \) to draw the direction of the curve.
• Incorporate \( f''(x) \) insights to adjust the curve for concavity. For example, early sections need to be sketched concave down before transitioning to a concave-up section, as the exercise guides.

Remember, a correctly sketched graph fulfills the continuity condition provided, having smooth and unbroken curves without jumps. This comprehensive approach ensures the graph accurately reflects the function's behavior.