Problem 7 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 1 through 12: (a) Find all critical points. (b) Find \(f^{\prime \prime} .\) Use the second derivative, wherever possible, to determine which critical points are local maxima and which are local minima. If the second derivative test fails or is inapplicable, explain why and use an alternative method for classifying the critical point. $$ f(x)=x^{4}+4 x^{3}+2 $$

Step-by-Step Solution

Answer

The critical points are \(x=0\) and \(x=-3\). Based on the second derivative test, \(x=-3\) is a local maximum. Following the first derivative test, the point \(x=0\) is also a local maximum.

Step 1: Find the first derivative

The first derivative of the function \( f(x)=x^{4}+4 x^{3}+2 \) is given by: \( f^{\prime}(x)=4x^{3}+12x^{2} \).

Step 2: Find the critical points

Setting the first derivative equal to 0 to find the critical points gives: \( 4x^{3}+12x^{2}=0 \). Factoring out \(4x^2\) gives \(x^2( x + 3 )=0\). Thus, the critical points are \(x=0\) and \(x=-3\).

Step 3: Find the second derivative

The second derivative of the function is given by \( f^{\prime \prime}(x)=12x^{2}+24x \).

Step 4: Use the second derivative to classify the critical points

Evaluate the second derivative at the critical points. At \(x = 0\), \(f^{''}(0) = 0\). Hence the second derivative test fails for this point. At \(x = -3\), \(f^{''}(-3) = 36 < 0 \). Therefore, according to the second derivative test, \(x = -3\) is a local maximum.

Step 5: Alternative method for classifying the failed point

Since the second derivative test failed for \(x = 0\), analyze the sign of the first derivative before and after this point. If the sign changes from positive to negative, then \(x = 0\) is a maximum, if it changes from negative to positive then it is a minimum. If it stays the same, it's a saddle point. In this case, \(f'(x)\) changes from positive to negative as \(x\) goes from negative to positive across zero, so \(x = 0\) is a local maximum.

Key Concepts

Second Derivative Test

The Second Derivative Test is a convenient method to determine whether a critical point is a local maximum or minimum. It involves taking the second derivative of the function, denoted as \(f''(x)\), and evaluating it at the critical points found when the first derivative is equal to zero. The rule of thumb is simple:

• If \(f''(x)>0\), the function has a local minimum at that point.
• If \(f''(x)<0\), it indicates a local maximum.
• If \(f''(x)=0\), the test is inconclusive, and other methods must be used to classify the critical point.

In the given exercise, the second derivative of the function \(f(x)=x^{4}+4x^{3}+2\) is \(f''(x)=12x^{2}+24x\). For the critical point at \(x=-3\), evaluating the second derivative yields a negative value \(f''(-3)=36 < 0\), hence indicating a local maximum at this point. However, at \(x=0\), the second derivative is zero, making the test inconclusive. Thus, it underscores the importance of an alternative method such as analyzing the first derivative's behavior around the critical point.

Local Maxima and Minima

Local maxima and minima, or the 'highs' and 'lows' of a function, are fundamental points where the function changes direction. Identifying them is essential in understanding the function's overall shape and behavior.

A local maximum is a point where the function's value is higher than all nearby points, while a local minimum is where it's lower. The first derivative test is another way to identify these points. By evaluating the first derivative's sign change around a critical point, you can determine if it's a maximum or a minimum:

First Derivative Test

• If the first derivative changes from positive to negative at a critical point, the function has a local maximum there.
• If it changes from negative to positive, there's a local minimum.
• If there's no sign change, the critical point might be a saddle point or an inflection point.

In the step-by-step solution, the second derivative test failed to classify the critical point at \(x=0\) since \(f''(0)=0\). Therefore, the first derivative test is used, revealing a sign change from positive to negative as we move through the critical point, indicating a local maximum at \(x=0\).

First Derivative Calculus

First derivative calculus is a cornerstone of understanding motion and change in functions. It involves taking the derivative of a given function to determine the rate at which the function values change with respect to its variable, typically noted as \(f'(x)\).

The derivative can provide information about the function's increasing or decreasing behavior and lead to the location of critical points—where the derivative is either zero or undefined. These are the points where the function could change its direction (turning points) or where the slope of the tangent line is flat.

• A positive derivative indicates an increasing function.
• A negative derivative suggests a decreasing function.
• A derivative equal to zero can signal a potential local maximum, minimum, or saddle point.

In the initial exercise, the first derivative \(f'(x)=4x^{3}+12x^{2}\) was found and set to zero to determine the critical points \(x=0\) and \(x=-3\). These points were then further analyzed with the second derivative test and the sign of the first derivative to classify them as local maxima or minima. The exercise demonstrates how calculus tools interconnect to provide a more comprehensive understanding of a function's characteristics.