Problem 25 - Calculus for Biology and Medicine

Question

Find \(c\) such that \(f^{\prime}(c)=0\) and determine whether \(f(x)\) has a local extremum at \(x=c .\) \(f(x)=(x+1)^{3}\)

Step-by-Step Solution

Answer

The derivative \(f'(x)\) equals zero at \(x = -1\), but \(f(x)\) has no local extremum at this point because the second derivative test is inconclusive and higher order derivative shows no sign change.

Step 1: Differentiate the Function

To find the derivative of the function, apply the power rule. The given function is \(f(x) = (x+1)^3\). The derivative, \(f'(x)\), is calculated as follows: \[ f'(x) = \frac{d}{dx}((x+1)^3) = 3(x+1)^2 \] This is because the derivative of \((x+1)^3\) is \(3(x+1)^2\).

Step 2: Set the Derivative to Zero

To find critical points where the function has a horizontal tangent line, set the derivative \(f'(x)\) to zero: \[ 3(x+1)^2 = 0 \] This equation tells us to solve for \(x\).

Step 3: Solve for Critical Points

Divide both sides of the equation by 3 to isolate \((x+1)^2\): \[ (x+1)^2 = 0 \] Taking the square root of both sides, we get: \[ x+1 = 0 \] Therefore, \(x = -1\). This is the critical point around which we will check for local extrema.

Step 4: Determine the Sign of the Second Derivative

To verify if there is a local extremum at \(x=-1\), we calculate the second derivative \(f''(x)\) and evaluate it at \(x=-1\). Differentiate \(f'(x) = 3(x+1)^2\) again: \[ f''(x) = 6(x+1) \] Substitute \(x = -1\) into \(f''(x)\): \[ f''(-1) = 6(-1 + 1) = 6(0) = 0 \] Since the second derivative at \(x = -1\) is zero, the second derivative test is inconclusive.

Step 5: Use Higher Order Derivatives or Consideration

Since the second derivative test was inconclusive, we use a higher order derivative. Differentiate \(f''(x) = 6(x+1)\) one more time to get the third derivative: \[ f'''(x) = 6 \] Since \(f'''(x) = 6\) is positive and not dependent on \(x\), there is no change across the sign of derivatives, confirming \(x = -1\) to be an inflection point rather than a minimum or maximum. Therefore, there is no local extremum at \(x = -1\).

Key Concepts

Derivative

When we talk about the derivative of a function, we are referring to a method to find how a function changes as its input changes. This is particularly useful in finding the slope of a function at any given point.
To calculate a derivative, we often use rules like the power rule, product rule, or chain rule. These help in differentiating various functions more efficiently without having to go through the definition of the derivative every time.
For instance, in our exercise, we have the function \(f(x) = (x+1)^3\). To find its derivative, we use the power rule, which states that the derivative of \(x^n\) is \(n \cdot x^{n-1}\). So, the derivative \(f'(x)\) becomes:

• \(f'(x) = 3(x+1)^2\)

This formula gives us the rate of change of the function \(f(x)\) with respect to \(x\). It allows us to identify where the function's slope is zero, indicating potential critical points.

Critical Points

Critical points are special values of \(x\) where the function's derivative is zero, or where the derivative does not exist. At these points, the function may have a horizontal tangent line, which can indicate a maximum, minimum, or a point of inflection.
For our exercise, we solve for \(x\) in the equation \(3(x+1)^2 = 0\). Solving this, we find:

• \(x = -1\)

This \(x\) value is our critical point. By understanding where the derivative equals zero, we can further analyze the function for potential local extrema.

Local Extrema

Local extrema refer to the smallest or largest values of a function in a small region around a critical point.
They are points where the function reaches a local minimum (valley) or a local maximum (peak) compared to its immediate surroundings. To determine whether a critical point is a local extremum, we usually use the first or second derivative test.
In our exercise, however, when we calculated the second derivative at the critical point \(x = -1\), we found \(f''(-1) = 0\). This result tells us that the second derivative test is inconclusive, meaning we don't immediately know whether \(x=-1\) is a minimum or maximum.
This often leads us to investigate further, either by using higher derivative tests or by scrutinizing the function's behavior around the critical point.

Second Derivative Test

The second derivative test is a handy method in calculus to decide whether a critical point is a local maximum, local minimum, or neither.
The second derivative, \(f''(x)\), gives us information about the concavity of the function:

• If \(f''(c) > 0\), the function is concave up, indicating a local minimum at \(x=c\).
• If \(f''(c) < 0\), the function is concave down, suggesting a local maximum at \(x=c\).
• If \(f''(c) = 0\), the test is inconclusive.

In the task at hand, since \(f''(-1)=0\), the test was inconclusive.
To resolve this, we explored the third derivative, \(f'''(x) = 6\), which was positive and led us to conclude that the critical point \(x=-1\) was not a local extremum, but rather an inflection point, explaining the smooth change of concavity without a peak or valley around that point.