Problem 3 - Calculus - AP Edition

Question

(a) Use both the first and second derivative tests to show that \(f(x)=3 x^{2}-6 x+1\) has a relative minimum at \(x=1\). (b) Use both the first and second derivative tests to show that \(f(x)=x^{3}-3 x+3\) has a relative minimum at \(x=1\) and a relative maximum at \(x=-1\).

Step-by-Step Solution

Answer

(a) The function has a relative minimum at \(x=1\). (b) The function has a relative minimum at \(x=1\) and a relative maximum at \(x=-1\).

Step 1: Calculate the first derivative for part (a)

We start by finding the first derivative of the function \(f(x) = 3x^2 - 6x + 1\). The derivative is calculated as follows: \(f'(x) = \frac{d}{dx}(3x^2) - \frac{d}{dx}(6x) + \frac{d}{dx}(1) = 6x - 6\).

Step 2: Find critical points for part (a)

Set the first derivative equal to zero to find critical points: \(6x - 6 = 0\). Solving for \(x\) gives \(x = 1\).

Step 3: Apply the second derivative test for part (a)

Calculate the second derivative: \(f''(x) = \frac{d}{dx}(6x - 6) = 6\). Since \(f''(1) = 6\), which is positive, \(x=1\) is a relative minimum by the second derivative test.

Step 4: Apply the first derivative test for part (a)

Evaluate \(f'(x)\) around \(x = 1\): for \(x = 0\), \(f'(0) = -6\), and for \(x = 2\), \(f'(2) = 6\). The sign changes from negative to positive at \(x = 1\), confirming a relative minimum.

Step 5: Calculate the first derivative for part (b)

Find the first derivative of the function \(f(x) = x^3 - 3x + 3\). Using the power rule, we get \(f'(x) = 3x^2 - 3\).

Step 6: Find critical points for part (b)

Set the first derivative equal to zero: \(3x^2 - 3 = 0\). Solve this equation: \(3(x^2 - 1) = 0\) leading to \(x^2 = 1\). Thus, the critical points are \(x = 1\) and \(x = -1\).

Step 7: Apply the second derivative test for part (b)

Calculate the second derivative: \(f''(x) = \frac{d}{dx}(3x^2 - 3) = 6x\). At \(x=1\), \(f''(1) = 6\) which is positive, indicating a relative minimum. At \(x=-1\), \(f''(-1) = -6\) which is negative, indicating a relative maximum.

Step 8: Apply the first derivative test for part (b)

Evaluate \(f'(x)\) around the critical points: for \(x = 0\), \(f'(0) = -3\) (between \(-1\) and \(1\)), for \(x = -2\), \(f'(-2) = 9\), and for \(x = 2\), \(f'(2) = 9\). The sign changes from positive to negative at \(x = -1\), indicating a relative maximum, and from negative to positive at \(x = 1\), indicating a relative minimum.

Key Concepts

First Derivative Test

The First Derivative Test is a useful tool in calculus to determine the nature of critical points within a function. To perform this test, you'll want to follow these steps:

• Find the first derivative: Compute the first derivative of the function, denoted as \( f'(x) \). This derivative represents the slope of the tangent line to the curve at any point on the graph.
• Determine critical points: Set \( f'(x) = 0 \) to find critical points. These are the points where the slope of the tangent line is zero, indicating possible relative maxima, minima, or points of inflection.
• Test intervals around critical points: To determine whether each critical point represents a relative minimum or maximum, substitute values from the intervals around these points into \( f'(x) \), checking for sign changes.

If the derivative changes from negative to positive at a critical point, then this indicates a relative minimum. Conversely, a change from positive to negative suggests a relative maximum, while no sign change signifies a saddle point.

Second Derivative Test

The Second Derivative Test provides an additional method to determine the nature of critical points for a given function by analyzing the behavior of the second derivative \( f''(x) \).
To apply this test, here's what you do:

• Find the second derivative: Calculate the second derivative from the first derivative, \( f'(x) \), obtaining \( f''(x) \).
• Evaluate at critical points: Substitute the critical points identified from the first derivative test into the second derivative.

Interpreting Results

• If \( f''(c) > 0 \) at a critical point \( c \), the function is concave up, indicating a relative minimum at that point.
• If \( f''(c) < 0 \), the function is concave down, indicating a relative maximum.
• If \( f''(c) = 0 \), the test is inconclusive, and the point may need further investigation with alternative methods.

This test provides a quicker route to determine extrema when the sign of \( f''(x) \) is decisive.

Critical Points

Critical points of a function are essential in understanding its graph's geometry. They occur where the first derivative \( f'(x) \) is zero or undefined, indicating potential peaks, troughs, or flat spots.

• Finding critical points: Solve \( f'(x) = 0 \) for \( x \) or determine where \( f'(x) \) does not exist. These values of \( x \) are your critical points.
• Importance of critical points: These points are where the function's behavior can change, marking intervals of increase or decrease or inflection.

Classification at Critical Points

By further applying derivative tests, you can classify these points:

• Relative Minima: Where the graph changes from decreasing to increasing.
• Relative Maxima: Where the graph changes from increasing to decreasing.
• Points of Inflection: Where the concavity changes, though these are not necessarily critical points unless \( f'(x) = 0 \).

Understanding where critical points are and what happens around them is crucial for graphing and analyzing functions in calculus.