Problem 35 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the point(s) of inflection of the graph of the function. $$ g(x)=2 x^{4}-8 x^{3}+12 x^{2}+12 x $$

Step-by-Step Solution

Answer

The function \( g(x) = 2x^4 - 8x^3 + 12x^2 + 12x \) has no inflection points.

Step 1: Calculate First Derivative

The first derivative \( g'(x) \) of the function \( g(x) = 2x^4 - 8x^3 + 12x^2 + 12x \) can be computed through applying the power rule of differentiation. The result is \( g'(x) = 8x^3 - 24x^2 + 24x + 12 \).

Step 2: Calculate Second Derivative

To find the potential inflection points, we now calculate the second derivative \( g''(x) \) again applying the power rule, which results in \( g''(x) = 24x^2 - 48x + 24 \).

Step 3: Find Possible Points of Inflection

Points of inflection are found where the second derivative changes its sign. To find these potential points, we set the second derivative equal to zero, \( 24x^2 - 48x + 24 = 0 \). This quadratic equation can be simplified through dividing all terms by 24, resulting in \( x^2 - 2x + 1 = 0 \). This is a perfect square, \( (x - 1)^2 = 0 \), which roots are x = 1 with multiplicity of 2.

Step 4: Determine if points are inflection points

To decide whether the obtained point x = 1 is inflection point, we have to check for the sign changes of the second derivative. Pick a number less than 1 (like 0) and greater than 1 (like 2), and plug them into the second derivative. The results \( g''(0) = 24 > 0 \) and \( g''(2) = 0 \) show that there is no sign change around x = 1. Hence, there are no inflection points for the function \( g(x) \).

Key Concepts

First Derivative

The first derivative, often represented as \( g'(x) \), is crucial in finding how a function behaves. This derivative tells us the slope or the rate of change of the original function at any given point. In terms of inflection points, while the first derivative itself does not directly indicate them, it is an essential step in the process. By determining \( g'(x) \), we lay the groundwork for calculating the second derivative, which provides the necessary insight into the concavity of the graph.

In our exercise, we applied the power rule to find \( g'(x) = 8x^3 - 24x^2 + 24x + 12 \). Each term in the original function \( g(x) = 2x^4 - 8x^3 + 12x^2 + 12x \) was differentiated individually:

• The first term, \( 2x^4 \), becomes \( 8x^3 \).
• The second term, \( -8x^3 \), becomes \( -24x^2 \).
• Next, \( 12x^2 \) turns into \( 24x \).
• Lastly, \( 12x \) differentiates to a constant 12.

Each step reflects how the power rule increases the efficiency of differentiation, making it a vital tool in calculus exercises.

The first derivative serves as a stepping stone towards more complex insights into our function's behavior.

Second Derivative

The second derivative, denoted \( g''(x) \), provides even deeper insights into the function's behavior. Specifically, it helps identify concavity and possible inflection points. An inflection point is where the graph changes concavity, going from concave up to concave down, or vice versa.

To find the second derivative in our example, we differentiate the first derivative that we calculated. Applying the power rule once more, we get \( g''(x) = 24x^2 - 48x + 24 \). Let's break down how this is done:

• The first term, \( 8x^3 \), transforms into \( 24x^2 \).
• The second term, \( -24x^2 \), becomes \( -48x \).
• The \( 24x \) differentiates into the constant \( 24 \).
• The constant 12 differentiates to zero and drops out of the equation.

Once we have \( g''(x) \), we set it to zero to find potential inflection points: \( 24x^2 - 48x + 24 = 0 \). Simplifying this equation leads us to the root \( x = 1 \). However, determining a true inflection point requires checking if the second derivative changes sign around \( x = 1 \). As shown, there is no sign change, indicating no inflection at this point.

Calculus Exercises

Calculus exercises, such as the one we've explored, serve as valuable tools in understanding core concepts like derivatives and inflection points. Inflection points are particularly intriguing as they indicate where a function's curvature changes. However, discovering these points requires a methodical approach through derivatives.

Here's a simple breakdown of the essential steps in solving for inflection points:

• First, calculate the first derivative of the function to understand its rate of change.
• Next, find the second derivative for insights into the function's concavity.
• Set the second derivative to zero to find potential inflection points and solve for these points.
• Finally, to confirm if these are true inflection points, test values around the potential points to check for concavity changes.

Through these exercises, not only do you sharpen your differentiation skills, but you also deepen your comprehension of fundamental calculus concepts. This structured approach helps build confidence, making complex mathematical processes more accessible and less daunting.