Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Concavity Theorem to determine where the given function is concave up and where it is concave down. Also find all inflection points. $$ T(t)=3 t^{3}-18 t $$

Step-by-Step Solution

Answer

The function is concave up for \( t > 0 \), concave down for \( t < 0 \), and has an inflection point at \((0,0)\).

Step 1: Calculate the First Derivative

To determine concavity, start by finding the first derivative of the function. The original function is given by \[ T(t) = 3t^3 - 18t. \] Differentiate the function with respect to \( t \):\[ T'(t) = \frac{d}{dt}(3t^3 - 18t) = 9t^2 - 18. \]

Step 2: Calculate the Second Derivative

Next, find the second derivative of the function, which will help identify concavity. Differentiate \( T'(t) = 9t^2 - 18 \) with respect to \( t \):\[ T''(t) = \frac{d}{dt}(9t^2 - 18) = 18t. \]

Step 3: Determine Signs of Second Derivative

To determine concavity, we look at where the second derivative is positive or negative.- Set \( T''(t) = 18t = 0 \) to find critical points: \( t = 0 \).- For \( t > 0 \), \( T''(t) = 18t > 0 \), so the function is concave up.- For \( t < 0 \), \( T''(t) = 18t < 0 \), so the function is concave down.

Step 4: Identify Inflection Points

Inflection points occur where the concavity changes, which is where \( T''(t) \) changes its sign.- We have transition at \( t = 0 \), since \( T''(t) \) changes from negative to positive.- Therefore, the inflection point is at \( t = 0 \). To find the function value at this point, \[ T(0) = 3(0)^3 - 18(0) = 0. \]- Thus, the inflection point is \((0, 0)\).

Key Concepts

First Derivative

The first derivative of a function measures the rate of change, or how the function's value changes with respect to an independent variable.
In this context, we have the function \[ T(t) = 3t^3 - 18t. \]By differentiating with respect to \( t \), we obtain the first derivative as \[ T'(t) = 9t^2 - 18. \]

• This derivative tells us how the graph of the function is sloped at each point.
• If \( T'(t) > 0 \), the function is increasing. If \( T'(t) < 0 \), it is decreasing.

However, our task is to study concavity, so we will take another derivative.

Second Derivative

The second derivative provides insight into the curvature of the function. In other words, it tells us about the concavity.
The original function's first derivative is\[ T'(t) = 9t^2 - 18. \]
Taking the derivative once more gives us the second derivative:\[ T''(t) = 18t. \]

• This derivative helps identify where the function is concave up or concave down.
• A positive \( T''(t) \) suggests the function is concave up.
• A negative \( T''(t) \) suggests it is concave down.

Thus, the second derivative is a crucial tool for analyzing the shape of the graph.

Inflection Points

Inflection points occur where a function's curve changes concavity. They happen where the second derivative is zero or undefined and changes sign.
In our example:

• We set \( T''(t) = 18t = 0 \).
• This gives \( t = 0 \) as a potential inflection point.

We check if the concavity indeed changes at \( t = 0 \):

• For \( t > 0, \ T''(t) = 18t > 0 \), indicating concave up.
• For \( t < 0, \ T''(t) = 18t < 0 \), indicating concave down.

Therefore, \( t = 0 \) is confirmed as an inflection point. At this point, \( T(0) = 0 \), making the inflection point \((0, 0)\).

Concave Up

When a function is concave up, its graph looks like a cup, curving upward. This occurs when the second derivative is positive.
In the case of \[ T''(t) = 18t, \]we observe that for:

• \( t > 0, \ T''(t) = 18t > 0, \) indicating that the function is concave up.

This means any values of \( t \) greater than zero result in the graph bending upwards. It's like the surface of a smile, where the slope is increasing.

Concave Down

The concept of concave down can be understood as when the function's graph shapes like an upside-down bowl. It occurs where the second derivative is negative.
For the second derivative\[ T''(t) = 18t, \]we determine:

• \( t < 0, \ T''(t) = 18t < 0, \) indicating that the function is concave down.

This means any values of \( t \) less than zero cause the graph to bend downwards. It's like an inverted smile, where the slope is decreasing across \( t<0 \). Thus, recognizing concavity helps visualize and understand the behavior of the function in different regions.