Problem 290 - Precalculus

Question

For the following exercises, use a calculator to approximate local minima and maxima or the global minimum and maximum. $$ f(x)=2 x^{3}-3 x-1 $$

Step-by-Step Solution

Answer

Critical points are at \( x = \pm\sqrt{\frac{1}{2}} \), approximated using a calculator to determine local or global minima/maxima.

Step 1: Find the Derivative

First, we need to find the derivative of the function \( f(x) = 2x^3 - 3x - 1 \) to determine the critical points. The derivative \( f'(x) \) is obtained by differentiating each term separately: \( f'(x) = 6x^2 - 3 \).

Step 2: Set Derivative Equal to Zero

We find the critical points by solving the equation \( f'(x) = 0 \). This gives \( 6x^2 - 3 = 0 \). Simplifying, we obtain \( x^2 = \frac{1}{2} \), which means \( x = \pm\sqrt{\frac{1}{2}} \).

Step 3: Evaluate Function at Critical Points

Substitute the critical points back into the original function \( f(x) \) to find their corresponding function values. These include \( f(\sqrt{\frac{1}{2}}) \) and \( f(-\sqrt{\frac{1}{2}}) \). Use a calculator to compute these values.

Step 4: Determine Nature of Critical Points

To determine whether each critical point is a local minimum or maximum, evaluate the second derivative \( f''(x) = 12x \). Then, substitute the critical points into \( f''(x) \). If \( f''(x) > 0 \), the point is a local minimum. If \( f''(x) < 0 \), the point is a local maximum.

Step 5: Evaluate End Behavior

Determine the behavior of the function as \( x \to \infty \) and \( x \to -\infty \). Since the leading term is \( 2x^3 \), the function increases without bound as \( x \to \infty \) and decreases without bound as \( x \to -\infty \).

Step 6: Compare Function Values

Compare the function values obtained at the critical points and the end behavior to establish which is the global minimum and maximum. Use a calculator to approximate these values precisely.

Key Concepts

Critical Points

In calculus, critical points are where the function's derivative is zero or undefined. These points are super important because they help us identify potential locations of local maxima or minima. In our exercise, we identified the critical points by setting the derivative of our given function, \( f(x) = 2x^3 - 3x - 1 \), equal to zero, which led us to solve \( 6x^2 - 3 = 0 \).

To find these points:

• Find the derivative, \( f'(x) = 6x^2 - 3 \).
• Set \( f'(x) = 0 \) to find potential critical points: \( x = \pm \sqrt{\frac{1}{2}} \).

These critical points are where the graph of the function might change direction, indicating possible peaks or valleys.

Derivative

The derivative of a function gives us the slope of the tangent line at any point on the function's graph. It indicates how fast the function's values are changing. For a polynomial, differentiating each term separately makes it easy to get the derivative.

For our function \( f(x) = 2x^3 - 3x - 1 \), the derivative is \( f'(x) = 6x^2 - 3 \). Here's how we obtained it:

• The derivative of \( 2x^3 \) is \( 6x^2 \) (using the power rule: multiply by the exponent and reduce the exponent by 1).
• The derivative of \(-3x\) is \(-3\).
• Since the derivative of a constant \(-1\) is 0, it disappears.

Understanding how to find the derivative is crucial in finding critical points and analyzing a function’s behavior.

Local Minima and Maxima

Local minima and maxima are the peaks and valleys in a function, where the function reaches its lowest or highest value locally. Once you have critical points, these are the spots to look! In our given problem:

Evaluate the function values at the critical points by plugging them back into the original function \( f(x) = 2x^3 - 3x - 1 \):

• Calculate \( f(\sqrt{\frac{1}{2}}) \) and \( f(-\sqrt{\frac{1}{2}}) \) using a calculator.

These values help determine if they are indeed local maxima or minima by comparing the function's behavior around these points. Essentially, they tell us where the graph crests or troughs.

Second Derivative Test

The second derivative test is a neat tool to determine the nature of critical points more precisely—whether they’re local minima, maxima, or something else. Once you have the critical points, evaluate the second derivative:

• For our function, the second derivative is \( f''(x) = 12x \).
• Substitute the critical points back into \( f''(x) \).

You interpret the result as follows:

• If \( f''(x) > 0 \) at a critical point, it is a local minimum (the function is concave up).
• If \( f''(x) < 0 \) at a critical point, it is a local maximum (the function is concave down).
• If \( f''(x) = 0 \), the test is inconclusive.

In our specific exercise, it allows us to decide which critical points are minima or maxima by examining the concavity of the function at these points.