Problem 68 - Thomas Calculus

Question

Sketch a smooth connected curve \(y=f(x)\) with $$ \begin{array}{ll}{f(-2)=8,} & {f^{\prime}(2)=f^{\prime}(-2)=0} \\ {f(0)=4,} & {f^{\prime}(x) < 0 \text { for }|x| < 2} \\ {f(2)=0,} & {f^{\prime \prime}(x) < 0 \text { for } x < 0} \\ {f^{\prime}(x) > 0 \text { for }|x| > 2,} & {f^{\prime \prime}(x) > 0 \quad \text { for } \quad x>0}\end{array} $$

Step-by-Step Solution

Answer

The curve has horizontal tangents at \(x = -2\) and \(x = 2\), decreases between \(-2\) and \(2\), and changes concavity around \(x = 0\).

Step 1: Identify Key Points

Based on the given conditions, identify the key points on the graph. We have:- \(f(-2) = 8 \): The curve passes through the point \((-2, 8)\).- \(f(0) = 4 \): The curve passes through the point \((0, 4)\).- \(f(2) = 0 \): The curve passes through the point \((2, 0)\).

Step 2: Determine Critical Points

Identify critical points where the derivative is zero:- \(f'(2) = 0 \): There is a horizontal tangent at \(x = 2\).- \(f'(-2) = 0 \): There is a horizontal tangent at \(x = -2\), suggesting extrema at these points.

Step 3: Analyze Increasing and Decreasing Intervals

From the conditions:- \(f'(x) < 0\) for \(|x| < 2\): The function is decreasing in the interval \((-2, 2)\).- \(f'(x) > 0\) for \(|x| > 2\): The function is increasing outside the interval \((-2, 2)\), i.e., for \(x < -2\) and \(x > 2\).

Step 4: Examine Concavity

Analyze the concavity using the second derivative:- \(f''(x) < 0\) for \(x < 0\): The function is concave down for \(x < 0\).- \(f''(x) > 0\) for \(x > 0\): The function is concave up for \(x > 0\).

Step 5: Sketch The Curve

Combine all the information to sketch the curve:- From \((-\infty, -2)\), the curve is increasing and concave down.- At \(x = -2\), there is a local maximum, passing through \((-2, 8)\).- From \((-2, 0)\), the curve is decreasing and concave down, passing through \((0, 4)\).- From \((0, 2)\), the curve continues to decrease but becomes concave up, ending at \((2, 0)\) with a horizontal tangent at 2.- From \((2, \infty)\), the curve begins increasing with concavity going upwards.

Key Concepts

Critical Points

Critical points are essential when sketching a curve because they often indicate where the function can turn or change its behavior. We find critical points by determining where the first derivative of the function equals zero or where the derivative does not exist. These points can suggest local maxima or minima, or sometimes saddle points.

In our problem, we see that the derivative, denoted as \(f'(x)\), is zero at two specific points: \(x = -2\) and \(x = 2\).
This implies that both points are critical points, potentially indicating local extrema. These are locations on the curve where the slope of the tangent is flat, or horizontal.
This means the function has horizontal tangents at these points. For a clear curve sketch, identifying where these horizontal tangents occur helps us determine where peaks and dips may appear.

Concavity

Concavity describes how the slope of a curve changes. It tells us whether a function is bending upwards like a cup or downwards like a frown. To find concavity, we use the second derivative, denoted as \(f''(x)\).

The sign of \(f''(x)\) helps determine concavity:

• If \(f''(x) > 0\), the curve is concave up (like a smile). The slope increases as \(x\) increases.
• If \(f''(x) < 0\), the curve is concave down (like a frown). The slope decreases as \(x\) increases.

In the exercise, we observe that:
- \(f''(x) < 0\) when \(x < 0\), indicating the function is concave down for all \(x\) less than zero.
- \(f''(x) > 0\) when \(x > 0\), indicating the function is concave up for \(x\) greater than zero.

This transition from concave down to concave up creates an inflection point at \(x = 0\). Though not strictly a point where the slope is zero, it marks a significant change in the curve's shape.

Increasing and Decreasing Functions

The behavior of a function regarding its increase or decrease is determined by its first derivative, \(f'(x)\). This tells us where the function is rising upwards or falling downwards as \(x\) moves.

We use the following rules to understand this behavior:

• If \(f'(x) > 0\), the function is increasing, meaning the curve ascends as \(x\) increases.
• If \(f'(x) < 0\), the function is decreasing, meaning the curve descends as \(x\) increases.

In our exercise:
- The function decreases for \(|x| < 2\), meaning it slopes downwards between \(-2\) and \(2\). This behavior results from \(f'(x) < 0\) in this interval.
- The function increases outside these bounds, i.e., for \(x < -2\) and \(x > 2\), where \(f'(x) > 0\).

Identifying these intervals helps outline the curve’s general behavior, giving us clear steps in the areas where the curve's movement changes direction.

Derivatives

Derivatives are fundamental for understanding and sketching curves. They provide insight into how a function changes and help determine crucial aspects such as slope, direction, and curvature.

We focus on two primary types of derivatives:

• The first derivative, \(f'(x)\), indicates the slope of the function at any point on the curve. Its value reveals where the function is increasing or decreasing, and where critical points occur.
• The second derivative, \(f''(x)\), gives us information about the concavity of the function. Its sign indicates whether the curve is bending upwards or downwards.

In this problem,
- We used \(f'(x)\) to locate critical points at \(x = -2\) and \(x = 2\), and understand the intervals of increase or decrease.
- We used \(f''(x)\) to determine the curve's concavity on either side of \(x = 0\).

Understanding derivatives thoroughly allows us to construct accurate curve sketches and interpret a function's behavior effectively.