Problem 14 - Single Variable Calculus

Question

Determine where the graph of the function is concave upward and where it is concave downward. Also, find all inflection points of the function. $$ h(x)=3 x^{4}+4 x^{3}+1 $$

Step-by-Step Solution

Answer

The graph of the function \(h(x) = 3x^4 + 4x^3 + 1\) is concave upward on all intervals (i.e., for all x-values), and there are no inflection points.

Step 1: Find the first derivative

To find the first derivative, take the derivative of h(x) with respect to x using the power rule. \(h'(x) = 12x^3 + 12x^2\)

Step 2: Find the second derivative

Take the derivative of the first derivative with respect to x to find the second derivative, using the power rule. \(h''(x) = 36x^2 + 24x\)

Step 3: Solve for when the second derivative is greater than zero

To determine where the graph is concave upward, we need to find the x-values for which the second derivative is greater than zero. \(36x^2 + 24x > 0\) Factor out the greatest common factor: \(12x(3x + 2) > 0\) Now, we'll find the critical points by setting the factors equal to zero: \(12x = 0 \Rightarrow x = 0\) \(3x + 2 = 0 \Rightarrow x = -\frac{2}{3}\) Using these critical points, create a sign chart and test intervals to determine when the second derivative is positive. On the sign chart, mark -2/3 and 0, test intervals (-∞, -2/3), (-2/3, 0), and (0, ∞): For interval (-∞, -2/3), let's test by plugging in x = -1: \(12(-1)(3(-1) + 2) = 12(-1)(1) > 0\), so concave up on interval \((-\infty, -\frac{2}{3})\). For interval (-2/3, 0), let's test by plugging in x = -1/2: \(12(-\frac{1}{2})(3(-\frac{1}{2}) + 2) = 12(-\frac{1}{2})(-0.5) > 0\), so concave up on interval \((-\frac{2}{3}, 0)\). For interval (0, ∞), let's test by plugging in x = 1: \(12(1)(3(1) + 2) = 12(1)(5) > 0\), so concave up on interval \((0, \infty)\). So the function is concave upward on all intervals.

Step 4: Solve for when the second derivative is less than zero

Since we have already determined that the second derivative is greater than zero on all intervals, there are no intervals where the graph is concave downward.

Step 5: Find inflection points

Inflection points occur where the second derivative is equal to zero or undefined. Since the second derivative is a polynomial, it has no points where it is undefined. Thus, inflection points occur where the second derivative is equal to zero. However, we already found there are no points where the second derivative is equal to zero when we determined concave upward intervals in Step 3. Thus, there are no inflection points for the function h(x). In conclusion, the graph of the function h(x) is concave upward on all intervals (i.e., for all x-values), and there are no inflection points.

Key Concepts

Concavity of Functions

Understanding the concavity of functions is crucial as it describes the curvature of the graph. A function is said to be concave upward if its graph bends like a cup (bowl-shaped), forming a U shape. Conversely, it is concave downward if it's dome-shaped, like an upside-down U. Concavity tells us about the acceleration of the function:

• If the second derivative is positive \(f''(x) > 0\), the function is concave upward.
• If the second derivative is negative \(f''(x) < 0\), the function is concave downward.

To determine the concavity of a polynomial function, we use its second derivative. By solving \(f''(x) > 0\) or \(f''(x) < 0\), we locate the x-values where the function changes its concavity.

Inflection Points

Inflection points are special points on the graph of a function where the concavity changes. When a function transitions from concave upward to concave downward or vice versa, these changes in curvature are marked by inflection points.

• Inflection points occur where the second derivative is equal to zero \(f''(x) = 0\) or undefined.
• These points show a change in the "direction" of the curve.

For polynomial functions, finding inflection points involves solving \(f''(x) = 0\). If an inflection point exists, it is located at these critical zero points. In our example function \(h(x) = 3x^{4} + 4x^{3} + 1\), we determined that there are no inflection points since the second derivative does not equal zero in our tested intervals.

Derivatives

Derivatives are foundational to calculus, representing rates of change. The first derivative of a function, denoted as \(f'(x)\), provides the slope of the tangent line at any point.

• The first derivative tells us whether the function is increasing or decreasing.
• The second derivative, \(f''(x)\), sheds light on the function's concavity and helps identify possible inflection points.

In our example, finding \(h'(x) = 12x^3 + 12x^2\) allows us to derive \(h''(x) = 36x^2 + 24x\). The second derivative is most useful here for determining the concavity of \(h(x)\) and evaluating inflection points as it further guides our analysis of the graph's shape.

Polynomial Functions

Polynomial functions are vital in mathematics due to their diverse applications. They consist of variables raised to whole number exponents and appear in the form of equations like \(h(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0\). With only integer exponents and coefficients, they provide predictable patterns that are ideal for analysis.

• These functions can be plotted on Cartesian planes, and their graphs display a smooth, continuous curve.
• The degree of a polynomial function (the highest power of x) affects its shape and features.

In the original exercise \(h(x) = 3x^{4} + 4x^{3} + 1\), this polynomial is of degree 4, indicating a quartic function. With even degree, it typically has end behaviors in the same direction, either up or down. Such functions make exploring concepts like derivatives, concavity, and inflection straightforward.