Problem 11 - 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. \(f(x)=x^{2}-5 x+3\)

Step-by-Step Solution

Answer

The critical point is \( x = \frac{5}{2} \), a local minimum. There are no inflection points.

Step 1: Find the First Derivative

To find the critical points, we first need the first derivative of the function. Given the function \( f(x) = x^2 - 5x + 3 \), the first derivative is obtained by differentiating with respect to \( x \). So, \( f'(x) = 2x - 5 \).

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. Set \( f'(x) = 2x - 5 = 0 \) and solve for \( x \). Thus, \( 2x = 5 \) which gives \( x = \frac{5}{2} \). There are no points where it's undefined since it's a linear polynomial.

Step 3: Calculate the Second Derivative

To determine the nature of the critical point and find inflection points, we need the second derivative. Differentiate \( f'(x) = 2x - 5 \) to get \( f''(x) = 2 \).

Step 4: Use the Second Derivative Test

The second derivative is a constant \( f''(x) = 2 \) which is greater than zero. This means the function is concave up, indicating the critical point at \( x = \frac{5}{2} \) is a local minimum.

Step 5: Identify Inflection Points

Inflection points occur where the second derivative changes sign. Since \( f''(x) = 2 \) never changes sign (it's a constant), there are no inflection points.

Step 6: Sketch the Graph

The quadratic function \( f(x) = x^2 - 5x + 3 \) is a parabola opening upwards. Plotting the graph confirms the critical point at \( x = \frac{5}{2} \) is indeed a local minimum. There are no points of inflection.

Key Concepts

First Derivative

The first derivative of a function helps us identify critical points, which are points on the function where the slope is zero or undefined. These points are crucial because they may indicate the presence of a local maximum or minimum. To find the first derivative of a function, you differentiate it with respect to the variable, typically denoted as \( x \). For the function \( f(x) = x^2 - 5x + 3 \), the first derivative is found using basic differentiation rules:

• Start with the power rule for differentiation: \( (x^n)' = nx^{n-1} \).
• Apply the rule to each term: from \( x^2 \) and \( -5x \), obtaining \( f'(x) = 2x - 5 \).

To locate critical points, set this first derivative equal to zero and solve for \( x \). In our example, this means solving \( 2x - 5 = 0 \), resulting in \( x = \frac{5}{2} \). There are no undefined points since the derivative is a simple linear equation.

Second Derivative

The second derivative provides insights into the concavity and points of inflection of a function. By taking the derivative of the first derivative, you identify the rate of change of the slope itself. For \( f(x) = x^2 - 5x + 3 \), the first derivative is \( f'(x) = 2x - 5 \). Differentiating this again yields:

• Apply the power rule to the linear term: The derivative of \( 2x \) is \( 2 \), and that of a constant is zero.
• This gives you \( f''(x) = 2 \), meaning the second derivative is a constant.

When the second derivative is constant and positive, it indicates the function is concave up throughout, and there are no changes in concavity indicating the lack of inflection points.

Inflection Points

Inflection points occur where a function changes concavity—from concave up to concave down, or vice versa. These points are where the second derivative equals zero or is undefined, and there is a sign change around these points. Examining the second derivative of our function, \( f''(x) = 2 \), reveals the following:

• Since it is a positive constant, the function is always concave up.
• There are no inflection points because the second derivative never crosses zero or changes sign.

Inflection points are crucial in understanding the "bending" nature of a function's graph, but in this particular function, there is a uniform bend throughout.

Local Maximum

A local maximum of a function occurs at a critical point where the function changes from increasing to decreasing. This change is marked by the first derivative's sign changing from positive to negative. Furthermore, the test for maximum requires the second derivative to be negative at that critical point. In our exercise:

• The sole critical point found was \( x = \frac{5}{2} \).
• However, the constant second derivative \( f''(x) = 2 \) shows the function is concave up, making a local maximum impossible at this point.

The absence of a local maximum aligns with the upward-opening nature of the quadratic function's parabola.

Local Minimum

A local minimum appears at a critical point where the function shifts from decreasing to increasing. At such points, the first derivative changes from negative to positive. Utilizing the second derivative test, a positive value here indicates the presence of a local minimum due to the concave-up condition of the parabola. From the exercise:

• The critical point \( x = \frac{5}{2} \) is tested with the second derivative \( f''(x) = 2 \).
• Since it is positive, this confirms a local minimum at \( x = \frac{5}{2} \).
• This aligns well with quadratic functions opening upwards, leading to the vertex being the lowest point. 

Thus, at \( x = \frac{5}{2} \), the function \( f(x) \) reaches its local minimum, consistent throughout as indicated by the graph and derivative tests.