Problem 67 - College Algebra

Question

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

Step-by-Step Solution

Answer

Local minimum at \(x = -\frac{1}{\sqrt{3}}\) with \(f(x) \approx -1.62\).

Step 1: Find the Derivative of the Function

To find local minima and maxima, we first need the derivative of the function. The given function is \(f(x) = x^3 - x - 1\). The derivative, \(f'(x)\), is calculated as follows:\[f'(x) = 3x^2 - 1\].

Step 2: Set the Derivative to Zero

To determine the critical points, solve for \(x\) where \(f'(x) = 0\). Thus, we solve:\[3x^2 - 1 = 0\]\[3x^2 = 1\]\[x^2 = \frac{1}{3}\]\[x = \pm \frac{1}{\sqrt{3}}\].

Step 3: Calculate Critical Points

The critical points are \(x = \frac{1}{\sqrt{3}}\) and \(x = -\frac{1}{\sqrt{3}}\). Substitute these values into the original function to calculate \(f(x)\): - For \(x = \frac{1}{\sqrt{3}}\), calculate \(f(\frac{1}{\sqrt{3}})\).- For \(x = -\frac{1}{\sqrt{3}}\), calculate \(f(-\frac{1}{\sqrt{3}})\).

Step 4: Use Calculator to Approximate Values

Using a calculator, approximate the values of the function at the critical points:- \(f(\frac{1}{\sqrt{3}})\) is approximately -0.38.- \(f(-\frac{1}{\sqrt{3}})\) is approximately -1.62.

Step 5: Determine Local Minima and Maxima

Comparing the function values at the critical points helps us identify:- \(x = -\frac{1}{\sqrt{3}}\) corresponds to a local minimum because \(f(-\frac{1}{\sqrt{3}}) = -1.62\) is lower.- \(x = \frac{1}{\sqrt{3}}\) likely corresponds to a local maximum because \(f(\frac{1}{\sqrt{3}}) = -0.38\) is higher.

Key Concepts

Critical Points

Critical points are where a function's derivative equals zero or is undefined. They are important in identifying the locations of local maxima, minima, or saddle points. In simpler terms, these points on a graph are where the slope is flat. For the function given, we found the derivative to be zero at certain points by solving the equation \( f'(x) = 0 \). These are your critical points, and they tell us where to focus our attention in determining the behavior of a function.
Finding critical points involves:

• Calculating the derivative of the function.
• Setting the derivative to zero or identifying where it doesn't exist.
• Solving for the variable to find the specific points.

By examining these values, we can decide where possible local extrema might occur, which leads us to the next point: derivative calculation.

Derivative Calculation

The derivative tells us how a function changes at any given point. It's a crucial tool for finding critical points because it shows where a function's slope is zero, indicating a potential extremum. For our function \( f(x) = x^3 - x - 1 \), the derivative is computed as follows: \[ f'(x) = 3x^2 - 1 \].
Here are the steps to calculate a derivative:

• Use basic derivative rules such as the power rule: \( x^n \rightarrow nx^{n-1} \).
• Differentiate each term separately.
• Simplify the resulting expression.

This derivative equation \( 3x^2 - 1 = 0 \) guides us in finding where the function flattens out, setting the stage for determining maxima and minima.

Function Approximation

Function approximation involves estimating function values at specific points, especially when an exact solution is not possible or practical. In calculus, it’s common to use calculators or software to approximate complex mathematical results.
Approximations are particularly useful:

• For confirming critical points quickly.
• When dealing with irrational numbers that can't be neatly expressed.
• To visualize or sketch behavior of a function over a particular range.

In our exercise, we found the critical points, then plugged these values into the original function to approximate the function's actual output using a calculator. Understanding how to approximate accurately helps in efficiently solving calculus problems.

Maxima and Minima

Maxima and minima are points where the function reaches its highest or lowest values locally. Around these points, the function changes direction.
Classifying maxima and minima involves:

• Identifying critical points using derivatives.
• Substituting these points back into the original function to find their corresponding values.
• Comparing values to determine whether they represent local maxima, local minima, or sometimes saddle points.

From our calculations, we determined that \( x = \frac{1}{\sqrt{3}} \) is likely a local maximum with a function value of approximately -0.38. Whereas \( x = -\frac{1}{\sqrt{3}} \) is a local minimum with a function value of about -1.62. Identifying these characteristics gives us greater insight into the function's overall behaviour, helping us understand its graphical shape better.