Problem 33 - Calculus Early Transcendentals

Question

Find the relative extrema using both first and second derivative tests. $$ f(x)=1+8 x-3 x^{2} $$

Step-by-Step Solution

Answer

The function has a local maximum at \(x = \frac{4}{3}\).

Step 1: Find the First Derivative

To find the relative extrema, we first need the first derivative of the function. Given the function \(f(x) = 1 + 8x - 3x^2\), we differentiate each term: \( f'(x) = 0 + 8 - 6x = 8 - 6x.\)

Step 2: Set the First Derivative to Zero

Find the critical points by setting the first derivative equal to zero and solving for \(x\):\(8 - 6x = 0\)Solve for \(x\): \(6x = 8\)\(x = \frac{4}{3}.\)

Step 3: Find the Second Derivative

Calculate the second derivative of the function to use in the second derivative test. The first derivative is \(f'(x) = 8 - 6x\), so the second derivative is:\(f''(x) = -6.\)

Step 4: Use the Second Derivative Test

Apply the second derivative test at the critical point. If \(f''(x)\) is positive, the function has a local minimum; if it is negative, the function has a local maximum. Evaluating at \(x = \frac{4}{3}\):Since \(f''\left(\frac{4}{3}\right) = -6 < 0\), the function has a local maximum at \(x = \frac{4}{3}.\)

Step 5: Use the First Derivative Test (Verification)

Double-check the critical point using the first derivative test. Examine \(f'(x) = 8 - 6x\) around \(x = \frac{4}{3}\):- Test an \(x\) slightly less than \( \frac{4}{3}\), say \(x = 1\): \(f'(1) = 8 - 6(1) = 2 > 0\), so \(f(x)\) is increasing.- Test an \(x\) slightly more than \( \frac{4}{3}\), say \(x = 2\): \(f'(2) = 8 - 6(2) = -4 < 0\), so \(f(x)\) is decreasing.Thus, \(x = \frac{4}{3}\) indeed corresponds to a local maximum.

Key Concepts

First Derivative Test

The First Derivative Test is a useful method for identifying relative extrema, such as local maxima and minima of a function. It involves examining the behavior of a function's slope before and after critical points. To apply this method, follow these simple steps:

• First, find the critical points by setting the first derivative of the given function equal to zero. Critical points are values of \( x \) where the derivative is either zero or undefined, indicating potential peaks or valleys.
• Second, test the sign of the first derivative, \( f'(x) \), by picking values slightly less than and slightly greater than each critical point. This will show whether the function is increasing or decreasing around these points.

If the function changes from increasing to decreasing, the critical point is a local maximum. If it changes from decreasing to increasing, that's a local minimum. In our original exercise, we identified a critical point at \( x = \frac{4}{3} \). Testing nearby points confirmed this is a local maximum because the function increases before \( x = \frac{4}{3} \) (where \( f'(x) > 0 \)) and decreases after (where \( f'(x) < 0 \)).
Applying this method clarifies the function's behavior around critical points.

Second Derivative Test

The Second Derivative Test offers another approach to identify relative extrema, providing a quicker method if the calculation goes smoothly. The test utilizes the function's concavity at the critical points, achieved by taking the second derivative.

• First, ensure that you've found the critical points by setting the first derivative to zero.
• Next, calculate the second derivative of the function. For the function \(f(x) = 1 + 8x - 3x^{2}\), the second derivative is \( f''(x) = -6 \).

By evaluating \( f''(x) \) at the critical points, we can determine the nature of the extrema. If \( f''(x) > 0 \), the function is concave up, indicating a local minimum at that point. If \( f''(x) < 0 \), the function is concave down, indicating a local maximum. In our exercise, \( f''\left(\frac{4}{3}\right) = -6 < 0 \), confirming there is a local maximum at \( x = \frac{4}{3} \). This test is particularly useful when the second derivative is simple to compute, offering a quick confirmation of the function’s behavior.

Critical Points

Critical points are essential in understanding the nature of a function's graph. They are basically the \( x \) values where the slope of the function changes, which can indicate a relative maximum, minimum, or a 'saddle point' where the function levels out temporarily.To find critical points:

• Derive the function to find its first derivative. This illustrates how the function's slope changes along the graph.
• Set the first derivative equal to zero and solve for \( x \). This gives you the critical points.

In the given problem, we determined that by differentiating \( f(x) = 1 + 8x - 3x^{2} \) to get \( f'(x) = 8 - 6x \). Setting \( 8 - 6x = 0 \) yields the critical point \( x = \frac{4}{3} \).
Critical points are important because they indicate where a function might change from increasing to decreasing or vice versa, which holds the key to unlocking the function's relative extrema using tests like the First and Second Derivative Tests.