Question

Consider the functions \(f(x)=\frac{1}{3} x^{3}-3 x^{2}+5 x+7\) and \(f^{\prime \prime}(x)=2 x-6 .\) Given that \(f\) is concave up where \(f^{\prime \prime}(x)>0\) and \(f\) is concave down where \(f^{\prime \prime}(x)<0,\) find where \(f\) is concave up and where \(f\) is concave down.

Short answer

The function is concave down on \((-\infty, 3)\) and concave up on \((3, \infty)\).

Step-by-step solution

Given Information

You are given the function: \[ f(x) = \frac{1}{3} x^{3} - 3 x^{2} + 5 x + 7 \] and its second derivative \[ f^{\text{''}}(x) = 2x - 6. \] The problem states that the function is concave up where \[ f^{\text{''}}(x) > 0 \] and concave down where \[ f^{\text{''}}(x) < 0. \]

Find the Critical Points of the Second Derivative

Set the second derivative equal to zero to find the critical points: \[ 2x - 6 = 0. \] Solve for \(x\): \[ 2x = 6 \] \[ x = 3. \]

Determine Intervals for Concavity

The critical point \(x=3\) divides the number line into two intervals: 1. \((-\infty, 3)\) 2. \((3, \infty)\)

Test the Intervals

To determine where the function is concave up or concave down, test a point in each interval: For \((-\infty, 3)\), choose a point such as \(x=0\): \[ f^{\text{''}}(0) = 2(0) - 6 = -6 \] Since \( f^{\text{''}}(0) < 0 \), the function is concave down on \((-\infty, 3)\).For \((3, \infty)\), choose a point such as \(x=4\): \[ f^{\text{''}}(4) = 2(4) - 6 = 2 \] Since \( f^{\text{''}}(4) > 0 \), the function is concave up on \((3, \infty)\).

Key concepts

Second Derivative Test

The second derivative test is a powerful tool in calculus. It helps us determine the concavity of a function. If we have a function's second derivative, we can find where the function is concave up or concave down.
The second derivative of a function, denoted as \( f^{''}(x) \), provides us with information on how the slope of the tangent line to the function is changing.
Here's how it works:

• If \( f^{''}(x) > 0 \) at a point, the function is concave up at that point.
• If \( f^{''}(x) < 0 \) at a point, the function is concave down at that point.

Using the second derivative test helps us understand the behavior of functions more deeply.

Concave Up

A function is considered to be concave up when its graph is curved upward, like a cup. This happens where the second derivative is greater than zero, \( f^{''}(x) > 0 \).
In other words, the slopes of the tangent lines to the function are increasing.
In our example, we found that the function \( f(x) = \frac{1}{3} x^{3} - 3 x^{2} + 5 x + 7 \) has a second derivative \( f^{''}(x) = 2x - 6 \). By testing intervals, we discovered that:

• The function is concave up on the interval \( (3, \, \infty) \).
• This means the graph curves upwards after \( x = 3 \).

Being able to identify concave up regions helps in understanding the increasing behavior of a function.

Concave Down

A function is said to be concave down when its graph curves downward, like an upside-down cup. This occurs where the second derivative is less than zero, \( f^{''}(x) < 0 \).
This means the slopes of the tangent lines to the function are decreasing.
In our case, using the second derivative \( f^{''}(x) = 2x - 6 \), we determined by interval testing that:

• The function is concave down on the interval \( (-\infty, \, 3) \).
• Thus, before \( x = 3 \), the graph of the function curves downward.

Identifying concave down regions is crucial for understanding the decreasing behavior of a function.

Critical Points

Critical points for concavity are points where the second derivative equals zero, \( f^{''}(x) = 0 \).
These points help us determine where the function's concavity changes.
In our exercise, the second derivative of the function \( f(x) \) is given by \( f^{''}(x) = 2x - 6 \). Setting this equal to zero gives us:

• \( 2x - 6 = 0 \)
• Solving this, we get \( x = 3 \).

So, \( x = 3 \) is a critical point. It divides the number line into intervals that we will test to see where the function is concave up or down.

Interval Testing

Interval testing involves using test points within the regions divided by critical points to determine concavity.
It goes like this:
After finding the critical point(s) – in our case \( x = 3 \) – we divide the number line into intervals:

• \( (-\infty, \, 3) \)
• \( (3, \, \infty) \)

We then choose any point within these intervals to test the sign of the second derivative.
1. Choose a point in \( (-\infty, \, 3) \), like \( x = 0 \):
\( f^{''}(0) = 2(0) - 6 = -6 \) (negative), so the function is concave down here.
2. Choose a point in \( (3, \, \infty) \), like \( x = 4 \):
\( f^{''}(4) = 2(4) - 6 = 2 \) (positive), so the function is concave up here.
Interval testing is straightforward but very informative for function analysis.