Problem 21 - Applied Calculus

Question

Find the inflection points of \(f(x)=x^{4}+x^{3}-3 x^{2}+2\).

Step-by-Step Solution

Answer

The inflection points are at \(x = -1\) and \(x = \frac{1}{2}\).

Step 1: Find the second derivative

Inflection points occur where the second derivative of the function changes sign. Start by differentiating the given function twice. The original function is:\[ f(x) = x^4 + x^3 - 3x^2 + 2 \]First, find the first derivative, \(f'(x)\):\[ f'(x) = 4x^3 + 3x^2 - 6x \]Then, find the second derivative, \(f''(x)\):\[ f''(x) = 12x^2 + 6x - 6 \]

Step 2: Set the second derivative to zero

To find potential inflection points, set the second derivative equal to zero and solve for \(x\):\[ 12x^2 + 6x - 6 = 0 \]Simplify by dividing the entire equation by 6:\[ 2x^2 + x - 1 = 0 \]

Step 3: Solve the quadratic equation

Solve the quadratic equation \(2x^2 + x - 1 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = 1\), and \(c = -1\).Plug in the values:\[x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}\]This simplifies to:\[x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4} = \frac{-1 \pm 3}{4}\]Thus, the solutions are:\[ x = \frac{2}{4} = \frac{1}{2} \] and \[ x = \frac{-4}{4} = -1 \]

Step 4: Check second derivative sign change

Verify a change in sign of \(f''(x)\) at each solution point to confirm they are inflection points.For \(x = \frac{1}{2}\) and \(x = -1\), check intervals:1. For \(x < -1\), choose \(x = -2\), then \(f''(-2) = 36 - 12 - 6 = 18 > 0\) (positive).2. For \(x = -1\), \(f''(-1) = 12(-1)^2 + 6(-1) - 6 = 12 - 6 - 6 = 0\).3. For \(-1 < x < \frac{1}{2}\), choose \(x = 0\), then \(f''(0) = -6\) (negative).4. For \(x = \frac{1}{2}\), \(f''(\frac{1}{2}) = 12(\frac{1}{2})^2 + 6(\frac{1}{2}) - 6 = 3 + 3 - 6 = 0\).5. For \(x > \frac{1}{2}\), choose \(x = 1\), then \(f''(1) = 12 + 6 - 6 = 12 > 0\) (positive).Both \(x = -1\) and \(x = \frac{1}{2}\) see a sign change in the second derivative, confirming them as inflection points.

Key Concepts

Second Derivative

The second derivative of a function helps to uncover valuable information about the shape and behavior of a graph. Specifically, it tells us about the curvature, or concavity, of a function.
- If the second derivative, denoted as \(f''(x)\), is positive at a certain interval, the graph is concave up, resembling a bowl facing upwards.- Conversely, if \(f''(x)\) is negative, the graph is concave down, like an upside-down bowl.
Inflection points occur where \(f''(x)\) changes sign. These points are crucial because they indicate where the graph switches from concave up to concave down, or vice versa. Finding an inflection point involves calculating the second derivative and determining where it equals zero or does not exist. For our exercise, the function was differentiated twice to find \(f''(x)=12x^2 + 6x - 6\). Identifying where this result changes sign reveals potential inflection points.

Quadratic Formula

A fundamental tool in solving quadratic equations is the quadratic formula. Given a quadratic equation of the form \(ax^2 + bx + c = 0\), the solutions for \(x\) can be found using:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula pulls roots directly from the coefficients \(a\), \(b\), and \(c\). It is very effective, especially when factoring is not straightforward.
In our exercise, the second derivative, \(2x^2 + x - 1 = 0\), was solved using this formula. Plugging into the formula, we arrive at the solutions, illustrating that this method is a powerful shortcut over manual factoring. Understanding how to use the quadratic formula opens doors to solving a variety of mathematical problems easily.

Sign Change

The concept of a sign change plays a pivotal role in determining the inflection points. Inflection points indicate where a graph changes its concavity, and this aspect relates directly to the sign of the second derivative.
- A change from positive to negative indicates the graph transitions from concave up to concave down.- A switch from negative to positive reveals a shift from being concave down to concave up.
In our specific problem, the computation showed a change in the sign of the second derivative at \(x = -1\) and \(x = \frac{1}{2}\). At these points, the sign of \(f''(x)\) shifts, confirming them as inflection points. These checks clarify important aspects of the graph's behavior and are essential for fully grasping the curve's properties.

Calculus

Calculus provides powerful techniques for understanding and analyzing functions. At its core lie derivatives, which describe how functions change. - The first derivative \(f'(x)\), informs us about the slope of the function or how steep it is at any given point.- The second derivative \(f''(x)\) takes this further by detailing the curvature, telling you if it is bending upwards or downwards.
The process of differentiating both first and second derivatives reveals vital information about a function's graph, including maxima, minima, and inflection points.
In this exercise, calculus guided us step-by-step to locate inflection points by differentiating twice to observe changes in signs of \(f''(x)\). This approach not only highlights the elegance of calculus but its utility in real-world applications, offering insights into trends and behaviors within data and mathematical models alike.