Problem 23 - Calculus for Biology and Medicine

Question

In Problems , 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

\( c = 0 \), no local extremum at \( x = 0 \).

Step 1: Find the derivative

First, we need to calculate the derivative of the function using power rules. The function is given as \( f(x) = x^3 \). The derivative \( f'(x) \) is found by differentiating: \( f'(x) = 3x^2 \).

Step 2: Set the derivative equal to zero

We need to find the value of \( c \) such that the derivative equals zero. Set \( f'(c) = 0 \): \( 3c^2 = 0 \).

Step 3: Solve for c

Solve the equation from step 2: \( 3c^2 = 0 \). Divide both sides by 3 to get \( c^2 = 0 \). Take the square root of both sides, resulting in \( c = 0 \).

Step 4: Determine the nature of the extremum using the second derivative test

The second derivative test helps determine the concavity at \( c \). First, find the second derivative \( f''(x) \) by differentiating \( f'(x) = 3x^2 \): \( f''(x) = 6x \). Evaluate the second derivative at \( c = 0 \): \( f''(0) = 6(0) = 0 \). Since \( f''(0) = 0 \), the test is inconclusive, indicating there is no local extremum at \( x = 0 \).

Key Concepts

Understanding Derivatives

The term "derivative" refers to the rate at which a function changes at any given point. It's a fundamental concept in calculus and is represented as \( f'(x) \) when you have a function \( f(x) \). In simpler terms, a derivative tells us how steep the graph of a function is at any point, or how the function "wiggles" around that point. For a function like \( f(x) = x^3 \), the derivative can be calculated using differentiation rules.

• The derivative \( f'(x) = 3x^2 \) was found by applying the power rule, which helps simplify calculations.
• Setting this derivative equal to zero helps in finding points where the function switches from increasing to decreasing, or vice versa.

In this case, setting \( 3x^2 = 0 \) and solving gives us \( c = 0 \). This point is crucial, as it might be where the function has a local extremum.

Applying the Second Derivative Test

The second derivative test is a method used to determine the nature of stationary points, or points where the first derivative equals zero. It checks whether these points are maxima, minima, or saddle points.

• To use the second derivative test, you first derive the second derivative of the function, \( f''(x) \).
• For \( f(x) = x^3 \), the second derivative is \( f''(x) = 6x \).

You then evaluate \( f''(x) \) at the point \( c \). The rule of thumb is:

• If \( f''(c) > 0 \), the point is a local minimum.
• If \( f''(c) < 0 \), it's a local maximum.
• If \( f''(c) = 0 \), the test is inconclusive.

For \( c = 0 \), \( f''(0) = 0 \), which means we can't definitively say whether there's a local extremum with this method. However, knowing how to apply this test is crucial for understanding more complex functions.

Using the Power Rule for Derivatives

The power rule is a quick method for finding the derivative of functions involving powers of \( x \). It's key for tackling polynomial functions efficiently. The basic form of the power rule states that if \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \).

• For \( f(x) = x^3 \), you apply the power rule to get \( f'(x) = 3x^2 \).
• This simplifies the process of differentiation, reducing complexity and saving time.

The power rule is essential for any calculus student because it forms the foundation of calculating derivatives for polynomial expressions. By mastering it, you can handle a wide range of function types with ease. Remember, it's all about bringing the power down as a coefficient, and then reducing the original power by one. This consistency makes it easier to work through more challenging problems later on.