Problem 23 - Calculus Early Transcendentals

Question

a. Find the critical points of the following functions on the domain or on the given interval. b. Use a graphing utility to determine whether each critical point corresponds to a local maximum, local minimum, or neither. $$f(x)=3 x^{2}-4 x+2$$

Step-by-Step Solution

Answer

Question: Identify the type of critical point (local maximum, local minimum, or neither) for the function $$f(x)=3x^2-4x+2$$ at $$x=\frac{2}{3}$$, and provide the corresponding \(f(x)\) value. Answer: The critical point $$x=\frac{2}{3}$$ corresponds to a local minimum. The corresponding \(f(x)\) value is $$f\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) + 2$$.

Step 1: Find the first derivative

To find the critical points, take the first derivative of the function with respect to \(x\): $$f'(x) = \frac{d}{dx}(3x^2 - 4x + 2)$$ Apply the power rule for each term: $$f'(x) = 6x - 4$$

Step 2: Find the critical points

Set the first derivative equal to zero and solve for \(x\): $$6x - 4= 0$$ $$x = \frac{2}{3}$$ Hence, the only critical point is \(x =\frac{2}{3}\).

Step 3: Find the second derivative

Calculate the second derivative of the function with respect to \(x\): $$f''(x) = \frac{d^2}{dx^2}(3x^2 - 4x + 2)$$ Apply the power rule: $$f''(x) = 6$$ Since the second derivative is a constant (6), it will always be positive.

Step 4: Use the second derivative test

The second derivative is positive at our critical point \(x=\frac{2}{3}\), so the point is a local minimum. We can also verify this using a graphing utility (such as Desmos, GeoGebra, or a graphing calculator) to confirm that the function has a local minimum at \(x=\frac{2}{3}\). To be more precise, the graph shows a local minimum at the point $$\left(\frac{2}{3}, f\left(\frac{2}{3}\right)\right)$$, corresponding to a value of $$f\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) + 2$$. So, based on our analysis and the use of a graphing utility, we can conclude that the critical point \(x=\frac{2}{3}\) corresponds to a local minimum.

Key Concepts

Critical Points

In calculus, critical points are key to understanding the behavior of a function. A critical point occurs where the first derivative of a function is either zero or undefined. These points are crucial because they can indicate turning points in the function's graph, which can be local maxima, local minima, or saddle points.

To determine the critical points of a function, follow these steps:

• Find the first derivative of the function.
• Set the first derivative equal to zero and solve for the variable. This gives potential critical points.
• Check where the first derivative is undefined to find additional critical points, if possible.

These locations may indicate changes in the direction of the function's graph. Therefore, finding the critical points helps in sketching the graph and analyzing the function's behavior.

First Derivative

The first derivative of a function, denoted as \(f'(x)\), provides valuable information about the slope of the function or its rate of change at any given point. It is a tool used to identify where a function is increasing, decreasing, or has a possible turning point.

In practical terms,

• The first derivative is positive when the function is increasing.
• The first derivative is negative when the function is decreasing.
• Where the first derivative equals zero, the function may have a local maximum, local minimum, or point of inflection.

To find the first derivative of a given function, apply differentiation rules such as the power rule, product rule, or chain rule, depending on the form of the function. Using these rules systematically allows you to uncover the critical points, which is fundamental for further analysis, like applying the second derivative test.

Second Derivative Test

The second derivative test provides a convenient way to classify each critical point found using the first derivative. This test utilizes the second derivative, \(f''(x)\), to determine the concavity of the function at a critical point.

Here's how the test works:

• If \(f''(x) > 0\) at a critical point, the graph is concave up, and the point is a local minimum.
• If \(f''(x) < 0\) at a critical point, the graph is concave down, and the point is a local maximum.
• If \(f''(x) = 0\), the test is inconclusive, and you may need other methods to determine the nature of the critical point.

By applying this test, you can quickly ascertain whether a critical point is a peak or a trough, aiding in a deeper understanding of the function’s graph and overall behavior.