Problem 30 - Calculus Early Transcendentals

Question

Use the given derivative to find all critical points of \(f\) and at each critical point determine whether a relative maximum, relative minimum, or neither occurs. Assume in each case that \(f\) is continuous everywhere. $$ f^{\prime}(x)=x^{4}\left(e^{x}-3\right) $$

Step-by-Step Solution

Answer

The critical points are \(x = 0\) (relative maximum) and \(x = \ln 3\) (relative minimum).

Step 1: Identify Critical Points

Critical points occur where the derivative is zero or undefined. Set the derivative to zero to find these points: \[ f'(x) = x^4 (e^x - 3) = 0 \]This results in two factors, so solve: 1. \( x^4 = 0 \) 2. \( e^x - 3 = 0 \)

Step 2: Solve for Critical Points

Solve the equations from Step 1:1. \( x^4 = 0 \rightarrow x = 0\)2. \( e^x = 3 \rightarrow x = \ln 3\)Thus, the critical points are \(x = 0\) and \(x = \ln 3\).

Step 3: Determine Nature of Critical Points using Second Derivative Test

To determine if these critical points are relative maxima or minima, we can use the second derivative test. First, find the second derivative: \[ f''(x) = 4x^3(e^x - 3) + x^4 e^x \]Evaluate the second derivative at each critical point.

Step 4: Check Critical Point at \(x = 0\)

Evaluate \(f''(0)\):\[ f''(0) = 4(0)^3(e^0 - 3) + (0)^4 e^0 = 0 \]The second derivative test is inconclusive at \(x=0\). Use the first derivative test around \(x=0\): Test x-values less than and greater than zero and observe the sign change (e.g., \(x = -1\) and \(x = 1\)).Since \(f'(x)\) changes from positive to negative, \(x=0\) is a relative maximum.

Step 5: Check Critical Point at \(x = \ln 3\)

Evaluate \(f''(\ln 3)\):\[ f''(\ln 3) = 4(\ln 3)^3(e^{\ln 3} - 3) + (\ln 3)^4 e^{\ln 3} \]Since \(e^{\ln 3} = 3\), it simplifies to:\[ f''(\ln 3) = (\ln 3)^4 imes 3 \]Since \((\ln 3)^4 > 0\), \(f''(\ln 3) > 0\), indicating a relative minimum at \(x=\ln 3\).

Key Concepts

Derivative

A derivative represents the rate at which a function changes as its input changes. It provides us with the slope of the tangent line to the curve of a function at any given point. In this exercise, the derivative is given by

• \( f'(x) = x^4 (e^x - 3) \).

This expression is critical because critical points occur where this derivative equals zero or becomes undefined. By focusing on the factors in the derivative, \( x^4 \) and \( e^x - 3 \), we set each factor to zero to find potential critical points. Because \( x^4 = 0 \) leads to \( x = 0 \) and \( e^x - 3 = 0 \) leads to \( x = \ln 3 \), these are our critical points. No part of this expression is undefined, as exponentials are defined everywhere, which simplifies our task.

Second Derivative Test

The Second Derivative Test helps determine the nature (minima or maxima) of critical points found using the first derivative. To apply this test, we find the second derivative of the function,

• \( f''(x) = 4x^3(e^x - 3) + x^4 e^x \).

We then evaluate this second derivative at each critical point.

• If \( f''(x) > 0 \), the point is a relative minimum.
• If \( f''(x) < 0 \), it's a relative maximum.
• If \( f''(x) = 0 \), further investigation is needed as the test is inconclusive.

In our example, the second derivative evaluated at \( x = \ln 3 \) is positive, showing a relative minimum, whereas at \( x = 0 \), it is zero, leading us to recheck using the first derivative.

Relative Maximum

A relative maximum is a point on the graph of the function where the function's value is higher than all nearby points. This occurs where the derivative changes from positive to negative.In our example, after checking using the first derivative test because the second derivative test was inconclusive at \( x = 0 \), we saw that the first derivative switches from positive to negative. Evaluating points slightly above and below zero confirmed this. Therefore, there is a relative maximum at \( x = 0 \). This technique is often used when the second derivative test isn't applicable.

Relative Minimum

A relative minimum is a point where the function's value is lower than that of nearby points. This happens when the derivative transitions from negative to positive, indicating a dip in the graph. In our exercise, evaluating the second derivative at \( x = \ln 3 \) gave a positive value, confirming a relative minimum. The positive result of \( f''(\ln 3) \) signifies that the graph is curved upwards around this point, forming a trough or valley. Finding such points helps in understanding the behavior of the function and is critical when sketching the curve or studying its real-world implications.