Problem 14 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find and classify all critical points. \(f(x)=x^{3}+x^{2}+x+1\)

Step-by-Step Solution

Answer

There are no critical points in the function \(f(x) = x^3 + x^2 + x + 1\).

Step 1: Find the derivative.

The derivative of \(f(x) = x^3 + x^2 + x + 1\) is calculated by applying the power rule to each term. That leads to \(f'(x) = 3x^2 + 2x + 1\).

Step 2: Find the critical points.

Setting the derivative equal to zero will give the critical points: \n0 = 3x^2 + 2x + 1. Solving this equation for \(x\) involves factorization or using the quadratic formula, in case factorization is not possible. This equation, however, can't be factored easily thus the quadratic formula seems to be the way to go:\n\(x = {-b \pm \sqrt{b^2 - 4ac}} \over 2a}\). Replace \(a\), \(b\), \(c\) with 3, 2, 1 respectively to find \(x\). However, the value inside the square root, \(b^2 - 4ac\), is less than zero, which means that there are no real roots for this equation and thus no critical points exist.

Step 3: Classify the critical points.

Since there are no real values of \(x\) where \(f'(x) = 0\), no critical points exist, and therefore, there is no need for classification with the second derivative test.

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function changes as its input changes. It is a fundamental concept used to understand how a function behaves. Imagine you are driving a car — the speedometer shows your current speed. This speed is similar to the derivative of your position with respect to time. In mathematical terms, if you have a function like \(f(x) = x^3 + x^2 + x + 1\), finding the derivative involves applying rules that help you determine how the function's values change as \(x\) changes.

In our example, we find the derivative \(f'(x)\) by applying the power rule. Calculating derivatives is crucial for identifying critical points, where a function's behavior changes significantly. These points help to locate the maximum, minimum, or inflection points on a graph. Understanding how to compute derivatives is an essential skill in calculus and provides insights into the function's behavior.

Quadratic Formula

When faced with a quadratic equation that can't be easily factored, we turn to the quadratic formula. This formula provides a reliable method for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). The quadratic formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In the problem we're looking at, you apply this formula to the derivative \(f'(x) = 3x^2 + 2x + 1\). By identifying \(a\), \(b\), and \(c\) in the equation as 3, 2, and 1 respectively, you can plug these values into the formula to find \(x\).

• This formula works by determining where the derivative equals zero, signifying potential critical points for the function.
• However, sometimes the result under the square root (the discriminant) is negative, which indicates there are no real solutions or critical points.

Understanding when and how to use the quadratic formula is key to solving various quadratic equations effectively in calculus.

Power Rule

The power rule is one of the simplest and most frequently used techniques in calculus for finding derivatives. It states that if you have a function in the form of \(f(x) = x^n\), the derivative \(f'(x)\) is \(nx^{n-1}\). This rule simplifies the process of differentiation for polynomial expressions, like the one in our example: \(f(x) = x^3 + x^2 + x + 1\).

The power rule is applied to each term in the function individually. Here's how it works step-by-step:

• For \(x^3\), applying the power rule gives \(3x^2\).
• For \(x^2\), it becomes \(2x\).
• For \(x\), the rule results in 1, since \(x^1\) becomes \(1 \cdot x^0 = 1\).
• A constant term like 1 disappears because its derivative is zero.

By using the power rule, you quickly and accurately find the derivative \(f'(x) = 3x^2 + 2x + 1\). This rule speeds up the differentiation process and is fundamental for solving more complex calculus problems efficiently.