Problem 8 - Calculus A New Horizon

Question

Give a complete graph of the polynomial, and label the coordinates of the stationary points and inflection points. Check your work with a graphing utility. $$3 x^{4}+4 x^{3}$$

Step-by-Step Solution

Answer

Stationary points: \((0,0), (-1,-1)\); Inflection points: \((0,0), (-\frac{2}{3}, -\frac{16}{9})\).

Step 1: Identify the Function

We are given the polynomial function \( f(x) = 3x^4 + 4x^3 \). We need to find its stationary points, inflection points, and subsequently graph it.

Step 2: Find the First Derivative

To find the stationary points, we need to calculate the first derivative of \( f(x) = 3x^4 + 4x^3 \). The derivative is \( f'(x) = 12x^3 + 12x^2 \).

Step 3: Solve for Stationary Points

Stationary points occur where the first derivative is zero. Solve \( 12x^3 + 12x^2 = 0 \). Factor out the greatest common factor: \( 12x^2(x + 1) = 0 \). Solve for \( x \), we find \( x = 0 \) and \( x = -1 \).

Step 4: Find the Second Derivative

To find the inflection points, compute the second derivative of \( f(x) = 3x^4 + 4x^3 \). The second derivative is \( f''(x) = 36x^2 + 24x \).

Step 5: Solve for Inflection Points

Inflection points occur where the second derivative is zero. Solve \( 36x^2 + 24x = 0 \). Factor: \( 12x(3x + 2) = 0 \). Solve for \( x \), giving \( x = 0 \) and \( x = -\frac{2}{3} \).

Step 6: Find Corresponding y-values

Determine the \( y \)-coordinates for the stationary and inflection points by substituting \( x \) back into the original function. \( f(0) = 0 \), \( f(-1) = -1 \), and \( f\left(-\frac{2}{3}\right) \approx -\frac{16}{9} \).

Step 7: Graph the Function

Using a graphing utility, graph \( f(x) = 3x^4 + 4x^3 \). Mark the stationary points at \((0,0)\) and \((-1,-1)\), and inflection points at \((0,0)\) and \((-\frac{2}{3}, -\frac{16}{9})\). Confirm that these points match graph characteristics.

Key Concepts

Stationary Points

In a polynomial function, stationary points play a crucial role in understanding the behavior of the graph. These points occur where the function's slope is zero, indicating a horizontal tangent. To find these points, we rely on the first derivative of the function.

The process involves:

• Calculating the first derivative of the given polynomial.
• Setting the derivative equal to zero and solving for the variable, here, "x".

For the polynomial function \(f(x) = 3x^4 + 4x^3\), the first derivative is \(f'(x) = 12x^3 + 12x^2\). We solve \(12x^3 + 12x^2 = 0\) to find the stationary points. Factoring gives us \(12x^2(x + 1) = 0\), leading to solutions \(x = 0\) and \(x = -1\).

These solutions mean that at \(x = 0\) and \(x = -1\), the polynomial has stationary points. It's essential to also calculate the corresponding \(y\)-values by plugging these \(x\)-values back into the original function: \(f(0) = 0\) and \(f(-1) = -1\). Thus, the stationary points for this function are \((0, 0)\) and \((-1, -1)\).

Inflection Points

Inflection points are where the curve of a function changes concavity, which means the graph shifts from being "bowl-shaped" (concave up) to "cap-shaped" (concave down), or vice versa. To find them, we need the second derivative of the function.

Here's how you locate inflection points:

• Calculate the second derivative of the polynomial.
• Set this derivative to zero and solve for \(x\).

For the polynomial \(f(x) = 3x^4 + 4x^3\), the second derivative is \(f''(x) = 36x^2 + 24x\). Solving \(36x^2 + 24x = 0\) gives \(12x(3x + 2) = 0\), leading to \(x = 0\) and \(x = -\frac{2}{3}\).

By finding corresponding \(y\)-values using the original function, we calculate \(f(0) = 0\) and \(f(-\frac{2}{3}) \approx -\frac{16}{9}\). Thus, inflection points occur at \((0, 0)\) and \((-\frac{2}{3}, -\frac{16}{9})\). Identifying these points helps in sketching a more accurate graph of the function over its domain.

Calculus Derivatives

Derivatives in calculus provide critical insights into the dynamics of a function's graph. They inform us about the slope and curvature of the function, which are fundamental in identifying stationary and inflection points.

Key aspects of understanding derivatives include:

• The first derivative \(f'(x)\) gives the slope of the tangent to the graph. Zero slopes point to potential local maxima or minima, also known as stationary points.
• The second derivative \(f''(x)\) reveals the concavity of the graph. Zero values of this derivative signal potential inflection points.

For our polynomial \(f(x) = 3x^4 + 4x^3\):- The first derivative is \(f'(x) = 12x^3 + 12x^2\). Identifying where this is zero helps pin down stationary points.- The second derivative is \(f''(x) = 36x^2 + 24x\). Solving this helps in locating inflection points.

These derivatives allow us to predict the function's behavior and sketch a graph with accuracy, providing essential characteristics such as peaks, valleys, and points of concavity change. Understanding derivatives is a building block for creating accurate representations of polynomial functions in calculus.