Problem 20 - The Calculus with Analytic Geometry

Question

\(f^{\prime \prime}(c)=0 ; f^{\prime}(c)=\frac{1}{2} ; f^{\prime \prime}(x)>0\) if \(xc\)

Step-by-Step Solution

Answer

Point c is an inflection point with a positive slope.

Step 1: Identify the Given Information

Given the following:- The second derivative at point c, \(f''(c) = 0\).- The first derivative at point c, \(f'(c) = \frac{1}{2}\).- The second derivative is positive before c, \(f''(x) > 0\) if \(x < c\).- The second derivative is negative after c, \(f''(x) < 0\) if \(x > c\).

Step 2: Determine the Nature of Point c

Since \(f''(c) = 0\) and the sign of \(f''(x)\) changes from positive to negative as \(x\) passes through c, this indicates that point c is an inflection point.

Step 3: Evaluate the First Derivative at Point c

Given that \(f'(c) = \frac{1}{2}\), we can see that the slope of the function at point c is positive.

Key Concepts

second derivative test

The second derivative test is a method used to determine the concavity of a function and identify possible points of local maxima, minima, or inflection points.
This test involves the second derivative of a function, denoted as \(f''(x)\).
Here are the key steps:
\(\textbf{1. Compute the second derivative:}\) Find the second derivative of the function, \(f''(x)\).
\(\textbf{2. Identify critical points:}\) Determine the points where the first derivative, \(f'(x)\), is zero or undefined. These are potential candidates for local maxima, minima, or inflection points.
\(\textbf{3. Evaluate the second derivative at critical points:}\) Substitute the critical points into the second derivative to assess concavity. If \(f''(x) > 0\), the function is concave up, indicating a local minimum; if \(f''(x) < 0\), the function is concave down, indicating a local maximum.
In the provided exercise, \(f''(c) = 0\), which means the second derivative test cannot definitively conclude if there is a local max or min at point c. Instead, we use a sign change in \(f''(x)\) to identify inflection points.

first derivative test

The first derivative test helps in identifying relative (local) maxima and minima of a function. It involves analyzing the sign of the first derivative, \(f'(x)\), around critical points.
Here’s how it works:
\(\textbf{1. Find the first derivative:}\) Compute \(f'(x)\), the first derivative of the function.
\(\textbf{2. Identify critical points:}\) Solve for points where \(f'(x) = 0\) or is undefined. These are potential positions for local maxima, minima, or inflection points.
\(\textbf{3. Analyze the sign of \(f'(x)\) around critical points:}\) Evaluate \(f'(x)\) just before and after the critical points. If \(f'(x)\) changes from positive to negative at a critical point, it indicates a local maximum. Conversely, if it changes from negative to positive, it suggests a local minimum.
In the given exercise, \(f'(c) = \frac{1}{2}\), signifying that the function has a positive slope at point c but does not affect the inflection point determination.

calculus inflection points

An inflection point is where the concavity of a function changes. At this point, the function does not necessarily have a local maximum or minimum. Instead, it shifts from being concave up to concave down, or vice versa.
Key steps to find inflection points:
\(\textbf{1. Second Derivative:}\) Calculate the second derivative, \(f''(x)\).
\(\textbf{2. Solve for zero:}\) Identify where \(f''(x) = 0\) or is undefined, indicating potential inflection points.
\(\textbf{3. Test intervals around the critical point:}\) Determine the sign of \(f''(x)\) before and after these points to confirm a change in concavity.
In the example, \(f''(c) = 0\) and \(f''(x)\) changes sign at x=c (positive for \(x < c\) and negative for \(x > c\)). Thus, c is an inflection point.

concavity

Concavity describes the curvature of a function’s graph, determined by the second derivative, \(f''(x)\).
There are two types of concavity:
\(\textbf{1. Concave up:}\) The graph bends upwards like a cup, and \(f''(x) > 0\). This indicates that the slope of the function is increasing.
\(\textbf{2. Concave down:}\) The graph bends downwards like an arch, and \(f''(x) < 0\). This means the slope of the function is decreasing.
Understanding concavity aids in sketching graphs and analyzing function behavior.
In the exercise, \(f''(x) > 0\) for \(x < c\) (concave up) and \(f''(x) < 0\) for \(x > c\) (concave down), signifying a concavity change at point c.