Problem 70 - College Algebra with Corequisite Support

Question

For the following exercises, use a calculator to approximate local minima and maxima or the global minimum and maximum. $$ f(x)=-x^{4}+3 x-2 $$

Step-by-Step Solution

Answer

The function has a local maximum at approximately \(x = 0.9086\) with value \(-0.2903\).

Step 1: Calculate the derivative of the function

First, we find the derivative of the function to locate the critical points, where minima and maxima can occur. The derivative of the function is given by: \[ f'(x) = \frac{d}{dx}(-x^4 + 3x - 2) = -4x^3 + 3 \]

Step 2: Solve for critical points

To determine the critical points, we need to set the derivative equal to zero and solve for \(x\): \[ -4x^3 + 3 = 0 \] Solving this equation, we get:\[ 4x^3 = 3 \] \[ x^3 = \frac{3}{4} \] \[ x = \sqrt[3]{\frac{3}{4}} \]

Step 3: Evaluate the function at the critical points

Using a calculator, evaluate \(f(x)\) at the critical point \(x = \sqrt[3]{\frac{3}{4}}\) to approximate the local extremum:Approximate \(x \approx 0.9086\) and calculate:\[ f(0.9086) = -(0.9086)^4 + 3 \times 0.9086 - 2 \approx -0.2903 \]

Step 4: Analyze endpoints if necessary

Since we are given only the function without interval constraints, we focus on the global and local extrema found using derivative methods. In general problems, check endpoints of the interval too, which are not specified here.

Step 5: Determine types of extremum

By examining the sign of the derivative around the critical point, we can infer the nature of the stationary point. However, we rely on numerical approximation here. Since \(f(x)\) decreases and increases around \(x = \sqrt[3]{\frac{3}{4}}\), this point corresponds to a local maximum.

Key Concepts

Derivative Calculation

Before exploring local minima and maxima, it's crucial to understand how to calculate the derivative of a function. The derivative provides information about the rate of change of a function, or in simpler terms, the slope of the tangent line at any point.
In the context of our exercise, we started with the function \[f(x) = -x^4 + 3x - 2\].
The derivative of this function using calculus is calculated as \[f'(x) = \frac{d}{dx}(-x^4 + 3x - 2) = -4x^3 + 3\].
This derivative provides us with information on how the function behaves and where it may have local minima or maxima.

Critical Points

Critical points are where the derivative of a function equals zero or does not exist. At these points, the function might reach a peak (maximum) or a valley (minimum).
To find them for our function \[f(x) = -x^4 + 3x - 2\],
we set the derivative equal to zero: \[-4x^3 + 3 = 0\].Solving this gives \[4x^3 = 3\],\[x^3 = \frac{3}{4}\],and finally,\[x = \sqrt[3]{\frac{3}{4}}\].
This calculation shows that the critical point is around \[x \approx 0.9086\].These critical points are potential spots for local maxima or minima, indicating where the function's slope is zero.

Numerical Approximation

Numerical approximation involves estimating the value of a function at certain points, especially when exact calculation is difficult. Calculations can involve rounding and using calculator power to handle complex numbers and functions.
Once the critical point \[x = \sqrt[3]{\frac{3}{4}}\]was identified, numerical approximation allowed us to estimate closely its value, around \[x \approx 0.9086\].Using this approximation, we calculated:\[f(0.9086) = -(0.9086)^4 + 3 \times 0.9086 - 2 \approx -0.2903\].
This helps in gauging the function's behavior around the critical point without analytical complexity.

Global Extremum

The global extremum refers to the highest or lowest point on the entire graph of a function. However, without endpoint constraints, as in this exercise, the task is to check if the found critical points also represent global extremum.In many scenarios, using derivative methods, we identify potential extrema at critical points. For \[x = \sqrt[3]{\frac{3}{4}}\]:1. We analyze changes in the function values around this critical point.2. Understanding whether there is a local maximum or minimum helps inform us of potential global behavior.Without specified intervals, we regard our local extrema check as tentative for global extremum presence. Indeed, checking intervals can further validate extremum points.