Problem 36 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the point(s) of inflection of the graph of the function. $$ f(x)=-4 x^{3}-8 x^{2}+32 $$

Step-by-Step Solution

Answer

The inflection point of the graph of the function \(f(x)=-4x^{3}-8x^{2}+32\) is at (\(\frac{2}{3}\), 29.63).

Step 1: Calculate the first derivative

The first derivative of the function \(f(x) = -4x^{3} - 8x^{2} + 32\) is found through standard rules of differentiation. \[f'(x) = \frac{d}{dx}[ -4x^{3} - 8x^{2} + 32] = -12x^{2}-16x \]

Step 2: Calculate the second derivative

The second derivative is the derivative of the first one. Find it by applying the normal rules of differentiation to the first derivative. \[f''(x) = \frac{d}{dx}[-12x^{2}-16x] = -24x - 16 \]

Step 3: Find the x-value(s) of the inflection point

You will get the x-coordinates of the inflection point by equating the second derivative to zero and solving for x. So, solve for x in the equation \(-24x - 16 = 0\). \[ x = -\frac{16}{-24} = \frac{2}{3} \]

Step 4: Find the y-value(s) of the inflection point

To get the y-coordinates of the inflection point, substitute x into the original function. \[f(\frac{2}{3}) = -4(\frac{2}{3})^3 - 8(\frac{2}{3})^2 + 32 \approx 29.63\] By applying these steps, the inflection point is at (\(\frac{2}{3}\), 29.63)

Key Concepts

Understanding the Second Derivative

The concept of the second derivative plays a crucial role in analyzing the behavior of functions, particularly when it comes to identifying inflection points.
In simple terms, the second derivative is the derivative of the derivative. While the first derivative gives us the slope or rate of change of a function, the second derivative provides information about the curvature or concavity of the graph of the function.

If the second derivative is positive, the graph is concave up, resembling a smiley face. On the other hand, if the second derivative is negative, the graph is concave down, like a frowny face.

• The sign of the second derivative at any point gives us an insight into whether the function is curving upwards or downwards.
• If it changes sign, it indicates the presence of an inflection point, which is where the curve changes concavity.

To find the second derivative, you apply the rules of differentiation to the first derivative. For the function provided, the second derivative was found to be \(f''(x) = -24x - 16\). This helps us to pinpoint the inflection point by analyzing where the derivative changes its sign.

Differentiation Explained

Differentiation is the process through which we find the derivative of a function. A derivative indicates how a function changes as its input changes.
This is a foundational tool in calculus that helps us understand various properties of functions like their slope and concavity.

• The first derivative gives us the slope of the function at any given point, which tells us whether the function is increasing or decreasing.
• By applying differentiation rules, like the power rule or product rule, we find derivatives easily for most functions.

In the exercise, the first step was to determine the derivative of \(f(x) = -4x^{3} - 8x^{2} + 32\), resulting in \(f'(x) = -12x^{2} -16x\).
This derivative provided the necessary information to further proceed with finding the second derivative, leading us closer to identifying the inflection point.

Identifying Critical Points and Inflection Points

Critical points are essential elements in the study of calculus, as they tell us where a function's rate of change dramatically shifts.
These points can define where a function reaches its maximum or minimum values, or where it alters its curvature.

• To find a critical point, look for where the first derivative equals zero or is undefined. This can indicate a potential maximum, minimum, or inflection point.
• Inflection points, in particular, occur where the second derivative is zero, implying a change in the concavity of the function.

In our exercise, by solving the equation \(-24x - 16 = 0\), we determined the x-coordinate of the inflection point to be \(x = \frac{2}{3}\).
Substituting this value back into the original function gives the y-coordinate, resulting in the inflection point at \(\left(\frac{2}{3}, 29.63\right)\).
This point shows where the function shifts from being concave down to concave up, illustrating a significant change in its behavior.