Problem 15 - The Calculus with Analytic Geometry

Question

Given \(f(x)=x^{3}+3 r x+5\), prove that (a) if \(r>0, f\) has no relative extrema; (b) if \(r<0, f\) has both a relative maximum value and a relative minimum value.

Step-by-Step Solution

Answer

(a) No relative extrema if \( r > 0 \). (b) One relative maximum and one relative minimum if \( r < 0 \).

Step 1 - Derive the First Derivative

To find the relative extrema, start by finding the first derivative of the function. For the given function, we have: \[ f(x) = x^3 + 3rx + 5 \]Use power rule and linearity to compute the first derivative: \[ f'(x) = 3x^2 + 3r \]

Step 2 - Set the First Derivative to Zero

To find critical points, solve for when the first derivative equals zero: \[ 3x^2 + 3r = 0 \]This simplifies to: \[ x^2 + r = 0 \]

Step 3 - Analyzing the Conditions

(a) If \( r > 0 \), the equation \( x^2 + r = 0 \) has no real solutions because \( x^2 \) is always non-negative and cannot equal a positive \( r \). Thus, the function has no critical points or relative extrema. (b) If \( r < 0 \), let \( r = -k \) where \( k > 0 \). Then, we have: \[ x^2 - k = 0 \]This simplifies to: \[ x^2 = k \]Resulting in: \[ x = \pm \sqrt{k} \]There are two critical points \( \sqrt{k} \) and \( -\sqrt{k} \).

Step 4 - Check the Second Derivative

To determine the nature of the critical points, evaluate the second derivative. For: \[ f'(x) = 3x^2 + 3r \]We derive again to get: \[ f''(x) = 6x \]At the critical points: \[ f''(\sqrt{k}) = 6\sqrt{k} > 0 \]\[ f''(-\sqrt{k}) = -6\sqrt{k} < 0 \]This means \( x = \sqrt{k} \) is a point of relative minimum and \( x = -\sqrt{k} \) is a point of relative maximum.

Key Concepts

First Derivative Test

The first derivative test is a key concept in calculus for identifying the relative extrema (maximum and minimum points) of a function. To use the first derivative test:

• First, derive the function to find the first derivative, noted as \( f'(x) \).
• Next, set the first derivative equal to zero to find the critical points (where the slope of the function is zero).
• Finally, analyze the sign changes of the first derivative around these critical points. If \( f'(x) \) changes from positive to negative through a critical point, you have a relative maximum. If \( f'(x) \) changes from negative to positive, there's a relative minimum.

In our exercise, we started by finding the derivative of \( f(x) = x^3 + 3rx + 5 \), which is \( f'(x) = 3x^2 + 3r \). Setting \( f'(x) = 0 \), we got \( x^2 + r = 0 \). This expression helps us determine the critical points based on whether \( r \) is positive or negative.

Second Derivative Test

The second derivative test provides another method to determine the nature of critical points. After finding the first derivative \( f'(x) \), the following steps apply:

• Calculate the second derivative \( f''(x) \) by deriving the first derivative again.
• Evaluate \( f''(x) \) at the critical points that were found from the first derivative.
• If \( f''(x) > 0 \) at a critical point, the point is a relative minimum because the function is concave up.
• If \( f''(x) < 0 \) at a critical point, the point is a relative maximum because the function is concave down.

From our exercise, we found the second derivative of \( f(x) = x^3 + 3rx + 5 \) to be \( f''(x) = 6x \). At the critical points \( x = \sqrt{k} \) and \( x = -\sqrt{k} \), it was determined that:

• At \( x = \sqrt{k} \), \( f''(x) = 6\sqrt{k} > 0 \), indicating a relative minimum.
• At \( x = -\sqrt{k} \), \( f''(x) = -6\sqrt{k} < 0 \), indicating a relative maximum.

Relative Maximum and Minimum

Relative maximum and minimum points are where a function temporarily reaches its highest and lowest values, respectively, within a given range. These points are not necessarily the highest or lowest values of the function for all possible values of \( x \), but only within the nearby values.

In our example, we were asked if \( f(x) = x^3 + 3rx + 5 \) has relative maximum and minimum points based on the value of \( r \).

• When \( r > 0 \), the equation \( x^2 + r = 0 \) has no real solutions, meaning the function has no relative extrema.
• When \( r < 0 \), substituting \( r = -k \) where \( k > 0 \), we found critical points \( \sqrt{k} \) and \( -\sqrt{k} \).

Using the second derivative test, we concluded that at \( \sqrt{k} \) there is a relative minimum and at \( -\sqrt{k} \) there is a relative maximum, thus proving that \( f(x) \) can indeed have relative extrema based on the value of \( r \).