Problem 17 - Applied Calculus

Question

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. $$f(x)=3 x^{4}-4 x^{3}+6$$

Step-by-Step Solution

Answer

Critical points: x=0 (max), x=1 (min); Inflection points: x=0, x=2/3.

Step 1: Find the First Derivative

To find the critical points, we first need to compute the first derivative of the function. Given \( f(x) = 3x^4 - 4x^3 + 6 \), we differentiate it with respect to \( x \). \[ f'(x) = \frac{d}{dx}(3x^4 - 4x^3 + 6) = 12x^3 - 12x^2 \]

Step 2: Solve for Critical Points

Critical points occur where the first derivative is zero or undefined. Here, the first derivative is a polynomial, so it is never undefined. Set it to zero and solve:\[ 12x^3 - 12x^2 = 0 \]Factor out \( 12x^2 \):\[ 12x^2(x - 1) = 0 \]This gives critical points at \( x = 0 \) and \( x = 1 \).

Step 3: Use the Second Derivative to Find Inflection Points

To find inflection points, we need the second derivative. Differentiate \( f'(x) \):\[ f''(x) = \frac{d}{dx}(12x^3 - 12x^2) = 36x^2 - 24x \]Set the second derivative equal to zero to find potential inflection points:\[ 36x^2 - 24x = 0 \]Factor out \( 12x \):\[ 12x(3x - 2) = 0 \]Solving, we find \( x = 0 \) and \( x = \frac{2}{3} \) as points to test for inflection.

Step 4: Test Concavity Change for Inflection Points

An inflection point requires a change in concavity, so test intervals around each candidate point \( x = 0 \) and \( x = \frac{2}{3} \) using the second derivative \( f''(x) = 36x^2 - 24x \):- For \( x < 0 \), such as \( x = -1 \), \( f''(-1) = 60 \) indicates \( f \) is concave up.- Check point \( x = 1/2 \), \( f''(1/2) = -6 \) indicates \( f \) is concave down.- For \( x > 1 \), such as \( x = 2 \), \( f''(2) = 72 \) indicates \( f \) is concave up.Thus, there is a change in concavity for \( x = 0 \) and \( x = \frac{2}{3} \), confirming inflection points there.

Step 5: Use a Graph to Classify Critical Points

Plot \( f(x) = 3x^4 - 4x^3 + 6 \) to visually identify characteristics of critical points:- At \( x = 0 \), the graph has a local maximum, as the slope changes from positive to negative.- At \( x = 1 \), the graph has a local minimum, as the slope changes from negative to positive.

Key Concepts

First Derivative

The first derivative of a function is a powerful tool in calculus, mainly used to find critical points where the function's growth rate changes. For the function given, \[f(x) = 3x^4 - 4x^3 + 6\], the first derivative is found by differentiating, resulting in \[f'(x) = 12x^3 - 12x^2\]. Critical points occur where this derivative is zero or undefined, and since this is a polynomial, it is defined everywhere. Therefore, setting \[12x^3 - 12x^2 = 0\] will help us find these points. Factoring gives \[12x^2(x-1) = 0\], indicating critical points at \(x = 0\) and \(x = 1\). Understanding where a function increases or decreases is crucial:

• When the first derivative is positive, the function is increasing.
• When it is negative, the function is decreasing.

Analyzing these intervals around critical points helps determine their nature, which can be classified further with the help of a graph.

Second Derivative

While the first derivative tells us about the slope of a function, the second derivative provides information on its curvature or concavity. For our function \[f(x) = 3x^4 - 4x^3 + 6\],the second derivative is computed by differentiating the first derivative: \[f''(x) = 36x^2 - 24x\]. This second derivative helps us pinpoint possible inflection points by setting \[36x^2 - 24x = 0\]. Factoring out \(12x\) gives \[12x(3x - 2) = 0\], indicating candidate points at \(x = 0\) and \(x = \frac{2}{3}\). Testing these points involves assessing the function's concavity:

• If the second derivative is positive, the graph is concave up, resembling a bowl shape.
• If negative, the graph is concave down, resembling an upside-down bowl.

This information offers insights into potential inflection points, where the concavity changes from up to down or vice versa.

Inflection Points

Inflection points are points on a curve where the concavity changes. To validate such points, not only we look for where the second derivative equals zero, but also a sign change in their intervals. **For our exercise:**- The candidate inflection points were found at \(x = 0\) and \(x = \frac{2}{3}\).By testing intervals around these points: - For example, using a test point, for **x < 0**, such as \(x = -1\), the second derivative \(f''(-1) = 60\), indicating the function is concave up. - At \(x = 1/2\), we find \(f''(1/2) = -6\), showing concave down. - Moving to \(x = 2\), the second derivative is \(f''(2) = 72\), concave up again.The change in concavity at these points confirms the presence of inflection points. Recognizing such points helps describe the overall shape of the graph, which is essential for understanding more complex function behaviors.