Problem 12 - Calculus with Concepts in Calculus

Question

Use the First Derivative Test to determine the relative extreme values (if any) of the function. $$ f(x)=x^{4}-8 x^{2}+1 $$

Step-by-Step Solution

Answer

Relative maximum at \( x = -2 \); relative minimum at \( x = 2 \).

Step 1: Find the First Derivative

To start, find the first derivative of the function \( f(x) = x^4 - 8x^2 + 1 \). To find the derivative, apply the power rule which states that \( \frac{d}{dx}[x^n] = n*x^{n-1} \). Thus, \( f'(x) = 4x^3 - 16x \).

Step 2: Set the First Derivative Equal to Zero

To find the critical points, we set the first derivative equal to zero: \( 4x^3 - 16x = 0 \). Factor the expression: \( 4x(x^2 - 4) = 0 \). This results in two equations: \( 4x = 0 \) and \( x^2 - 4 = 0 \).

Step 3: Solve for Critical Points

Solve the equations obtained from the previous step.\( 4x = 0 \) gives \( x = 0 \). \( x^2 - 4 = 0 \) can be factored as \( (x - 2)(x + 2) = 0 \), thus \( x = 2 \) and \( x = -2 \). Therefore, the critical points are \( x = 0, -2, 2 \).

Step 4: Test Intervals Around Critical Points

Use the first derivative test by choosing test points in the intervals created by the critical points \(-\infty, -2, 0, 2, \infty\). Evaluate the sign of \( f'(x) \) in each interval:- Interval \((-\infty, -2)\), test at \( x = -3 \): \( f'(-3) > 0 \)- Interval \((-2, 0)\), test at \( x = -1 \): \( f'(-1) < 0 \)- Interval \((0, 2)\), test at \( x = 1 \): \( f'(1) < 0 \)- Interval \((2, \infty)\), test at \( x = 3 \): \( f'(3) > 0 \)

Step 5: Determine Relative Extrema

Using the sign changes of \( f'(x) \):- At \( x = -2 \): Since \( f'(x) \) changes from positive to negative, a relative maximum occurs.- At \( x = 0 \): No sign change in \( f'(x) \) (negative to negative), no extrema.- At \( x = 2 \): Since \( f'(x) \) changes from negative to positive, a relative minimum occurs.

Key Concepts

Calculus

Calculus is a field of mathematics that studies how things change. It's very useful for understanding the behavior of functions and graphs. Calculus can be divided into two main areas: differential calculus and integral calculus.

Differential calculus focuses on the concept of the derivative. The derivative of a function measures how the function's output changes as the input changes. In simple terms, it helps you understand the slope of a curve at any point. This is crucial for various real-life applications, such as finding rates of change and understanding motion.

When you calculate a derivative, you're determining the rate at which a function is changing at any given point. This makes derivatives essential for identifying trends and behaviors in mathematical models. In this exercise, finding the first derivative is key to analyzing critical points and relative extrema.

Critical Points

Critical points are points on a graph where the derivative is zero or undefined. At these points, the function may change direction, which is why they're so important in calculus.

To find critical points, you take the first derivative of the function and set it equal to zero. This step tells you where the slope of the function's graph is horizontal. In the example exercise, the function is given as \( f(x) = x^4 - 8x^2 + 1 \). The first derivative is \( f'(x) = 4x^3 - 16x \). Setting the first derivative equal to zero gives \( 4x(x^2 - 4) = 0 \). Solving this equation helps us find the critical points: \( x = 0, -2, \) and \( 2 \).

These values are essential for identifying where the function might have highs or lows.

Relative Extrema

Relative extrema are points on a graph where a function reaches a relative maximum or minimum. These points occur at critical points in the function.

To determine the relative extrema from critical points, we use the First Derivative Test. This test involves checking the sign of the first derivative before and after the critical points. In the exercise, testing intervals around the critical points \(-2, 0, \) and \( 2 \) reveals the nature of these points:

• At \( x = -2 \), \( f'(x) \) changes from positive to negative, indicating a relative maximum.


• At \( x = 0 \), \( f'(x) \) does not change sign (it stays negative), so there is no relative extrema.


• At \( x = 2 \), \( f'(x) \) changes from negative to positive, indicating a relative minimum.

Understanding relative extrema helps in sketching graphs and predicting outcomes.

Power Rule

In calculus, the power rule is a basic technique used to find the derivative of a function with a power of \( x \). This rule states that if you have a function \( f(x) = x^n \), its derivative \( f'(x) \) is \( nx^{n-1} \).

The power rule makes differentiation straightforward, especially for polynomial functions. It's applied to each term individually. In the exercise, to find the derivative of \( f(x) = x^4 - 8x^2 + 1 \), the power rule is applied to get \( f'(x) = 4x^3 - 16x \). Each term's exponent is reduced by one, and the original exponent becomes a multiplier.

The power rule is simple but powerful, enabling quick and efficient differentiation of many functions you'll encounter in calculus.