Problem 23 - Calculus for Biology and Medicine

Question

Find \(c\) such that \(f^{\prime}(c)=0\) and determine whether \(f(x)\) has a local extremum at \(x=c .\) \(f(x)=x^{3}\)

Step-by-Step Solution

Answer

The critical point is at \( c = 0 \) with no local extremum, as \( f(x) = x^3 \) has an inflection point at \( x = 0 \).

Step 1: Find the derivative

The first step is to find the derivative of the function \( f(x) = x^3 \). The derivative, \( f'(x) \), is calculated using the power rule, which states that \( \frac{d}{dx}[x^n] = nx^{n-1} \). Applying this to \( x^3 \), we get: \[ f'(x) = 3x^2. \]

Step 2: Set derivative to zero

Next, to find the critical points, we set the derivative equal to zero: \[ 3x^2 = 0. \] This equation simplifies to \( x^2 = 0 \). Solving for \( x \), we find that \( x = 0 \). Thus, \( c = 0 \) is the critical point where the derivative is zero.

Step 3: Determine the nature of the critical point

To assess whether there is a local extremum at \( x = c \), we examine the second derivative, \( f''(x) \). First, we compute the second derivative: \[ f''(x) = \frac{d}{dx}[3x^2] = 6x. \] Substitute \( x = 0 \) into the second derivative: \[ f''(0) = 6(0) = 0. \] Since \( f''(c) = 0 \), we cannot conclude whether there is a local extremum using the second derivative test. Therefore, we examine the original function's behavior. For \( f(x) = x^3 \), it is known that \( x = 0 \) is an inflection point, not a local extremum.

Key Concepts

Power Rule

In calculus, the power rule is a simple and essential tool used to take derivatives of functions involving powers of a variable.
It's expressed as \( \frac{d}{dx}[x^n] = nx^{n-1} \), where \( n \) represents the power.
This means, to differentiate a term like \( x^3 \), you multiply the exponent by the coefficient (here, it's 1), and then reduce the exponent by one.
By applying this rule to \( x^3 \), we get the derivative \( 3x^2 \).

• It helps find the rate at which the function changes.
• Commonly used for polynomial functions.
• Simplifies the differentiation process significantly.

Understanding the power rule is key to solving many calculus problems, as it allows you to identify how individual terms in polynomials impact the slope of the curve at any point along a graph.
In our original exercise, applying the power rule gives us the derivative, which we use to find critical points.

Critical Points

Critical points of a function occur where its derivative is zero or undefined.
For polynomial functions, you'll mostly need to set the derivative equal to zero to find the critical points.

• The solution to the equation \( f'(x) = 0 \) gives these points.
• Each critical point is a candidate for a local extremum or an inflection point.

For our function \( f(x) = x^3 \), the derivative \( 3x^2 = 0 \) simplifies quickly to yield \( x = 0 \) as a critical point.
This indicates a point where the slope of the tangent to the curve is zero, meaning the graph is flat at this point.

Second Derivative Test

The second derivative test helps determine the nature of critical points found in the first derivative step.
If \( f''(x) \) at a critical point is non-zero, we can conclude if it's a local minimum or maximum:

• If \( f''(c) > 0 \), the function has a local minimum at \( x = c \).
• If \( f''(c) < 0 \), the function has a local maximum at \( x = c \).

In our problem, the second derivative is \( 6x \).
At \( x = 0 \), \( f''(0) = 0 \), meaning the test is inconclusive.
This tells us more analysis is required to fully understand the behavior of \( f(x) \) at that point.

Inflection Point

An inflection point occurs where the curvature of the function changes from concave up to concave down, or vice versa.
It's where the second derivative changes sign.

• Characteristics: Not a maximum or minimum.
• Important for understanding the shape of the graph.
• Often occurs where fourth or higher derivatives provide no clarity.

For \( f(x) = x^3 \), even though the second derivative test did not determine a local extremum at \( x = 0 \), \( x = 0 \) is an inflection point.
This is because the graph of \( f(x) \) changes its direction of curvature, making it a key feature in the graph's overall shape.