Problem 292 - Precalculus

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 local maximum occurs at \(x \approx 0.908\) with a value of \(f(x) \approx -0.681\).

Step 1: Identify the Function

First, look at the function given: \[ f(x) = -x^4 + 3x - 2 \]This is a polynomial function of degree 4, so the graph will have certain characteristic shapes such as turning points or inflection points.

Step 2: Take the Derivative

To find the local minima and maxima, we start by taking the derivative of the function. This helps in finding the critical points.The derivative of \( f(x) \) with respect to \( x \)is given by:\[f'(x) = \frac{d}{dx}(-x^4 + 3x - 2) = -4x^3 + 3\]

Step 3: Solve for Critical Points

Set the derivative equal to zero and solve for \( x \)to find the critical points:\[-4x^3 + 3 = 0\]Rearrange:\[4x^3 = 3\]Divide by 4:\[x^3 = \frac{3}{4}\]Take the cube root of both sides:\[x = \sqrt[3]{\frac{3}{4}}\]

Step 4: Approximate Critical Point

Using a calculator, approximate the cube root to find the x-coordinate of the critical point:\[x \approx 0.908\]

Step 5: Analyze Second Derivative

To determine if this point is a local minimum or maximum, find the second derivative:\[f''(x) = \frac{d}{dx}(-4x^3 + 3) = -12x^2\]Evaluate \( f''(x) \)at \( x \approx 0.908 \):\[f''(0.908) = -12(0.908)^2 \approx -9.924\]Since \( f''(x) < 0 \), this point is a local maximum.

Step 6: Evaluate Function Value

Using the x-coordinate, calculate the function's value at the local maximum:\[f(0.908) = -(0.908)^4 + 3(0.908) - 2 \approx -0.681\]

Key Concepts

Local Maxima and Minima

Local maxima and minima are essential to understanding the behavior of polynomial functions. These points are where a function reaches its highest or lowest value in a particular region.
Determining the local maxima and minima involves finding the turning points of a function. These are places where the graph of the function changes direction.

• **Local Maximum:** A point where the function value is higher than any nearby points.
• **Local Minimum:** A point where the function value is lower than any nearby points.

Finding these points requires calculus techniques, such as taking derivatives and applying tests that identify the nature of these points.
This understanding helps in predicting and analyzing the behavior patterns of the function as x changes.

Critical Points

Critical points are where the most significant changes occur in the function's behavior.
These points are crucial as they indicate where the slope of the tangent to the curve is zero or undefined.
To find critical points:

• First, take the derivative of the function.
• Set the derivative equal to zero and solve for the variable. This gives potential critical points.

The original exercise identifies critical points by setting the derivative of \[-4x^3 + 3\] equal to zero,
leading to the solution \[x = \sqrt[3]{\frac{3}{4}}\].
This point indicates where the function's slope is zero, a necessary condition for it to be a local maximum or minimum.

Derivative

A derivative represents the rate at which a function changes with respect to its variable. In simpler terms, it tells us how steep a curve is at any point.
For polynomial functions, calculating the derivative is straightforward and follows set rules:

• Reduce the power of x by one and multiply it by the original power.
• Sum these individual derivatives for each term.

In the given exercise, the function \[-x^4 + 3x - 2\]
has a derivative of \[-4x^3 + 3\].
This derivative helps identify the critical points where the slope of the function is zero. It provides essential insight into the function's increasing or decreasing behavior.

Second Derivative

The second derivative is the derivative of the derivative of a function.
This concept helps determine the concavity of the graph at a particular point, indicating whether the point is a local maximum or minimum.

• **Concave Up:** If the second derivative is positive, the graph is shaped like a cup.
• **Concave Down:** If the second derivative is negative, the graph is shaped like a cap.

In the problem, evaluating the second derivative \[-12x^2\] at \[x \approx 0.908\]
yields a negative result, indicating a concave down shape, thus confirming a local maximum.
Understanding the second derivative is crucial for detailed analysis of the function's shape and the nature of its critical points.