Problem 24 - Essential Calculus Early Transcendentals

Question

\(23-36=\) Find the critical numbers of the function. $$f(x)=x^{3}+6 x^{2}-15 x$$

Step-by-Step Solution

Answer

The critical numbers are \(x = -5\) and \(x = 1\).

Step 1: Find the first derivative

To find critical numbers, we start by finding the first derivative of the function. The given function is \(f(x) = x^3 + 6x^2 - 15x\). Using the power rule, the derivative is:\[f'(x) = 3x^2 + 12x - 15.\]

Step 2: Set the first derivative to zero

Critical numbers occur where the first derivative is zero or undefined. Since our derivative is a polynomial, it is defined everywhere. Therefore, we set \(f'(x)\) to zero:\[3x^2 + 12x - 15 = 0.\]

Step 3: Solve the quadratic equation

Solve the quadratic equation \(3x^2 + 12x - 15 = 0\). First, simplify by dividing the entire equation by 3:\[x^2 + 4x - 5 = 0.\]Next, factor the quadratic:\[(x + 5)(x - 1) = 0.\]The solutions to this equation are \(x = -5\) and \(x = 1\).

Step 4: Verify the solutions

Since both solutions from the factored form are valid, \(x = -5\) and \(x = 1\) are critical numbers of the function \(f(x)\). No solutions were lost during simplification or factoring.

Key Concepts

First Derivative

In calculus, the first derivative of a function represents the rate at which the function's value changes with respect to change in input value, often denoted as \( f'(x) \). The process of finding derivatives is called differentiation. To find the first derivative of a polynomial function like \( f(x) = x^3 + 6x^2 - 15x \), we use basic rules of differentiation, such as the power rule. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \), where \( n \) represents any real number.

Applying the power rule to each term in the given function:

• The derivative of \( x^3 \) is \( 3x^2 \).
• The derivative of \( 6x^2 \) is \( 12x \).
• The derivative of \( -15x \) is \( -15 \).

By summing these results, we obtain the first derivative: \( f'(x) = 3x^2 + 12x - 15 \). This derivative function provides critical information about where the original function's rate of change is zero or changes direction.

Polynomial Function

A polynomial function is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. An example of a polynomial function is \( f(x) = x^3 + 6x^2 - 15x \).

Polynomial functions are characterized by:

• Everything in it is either a constant, a variable raised to a non-negative integer power, or a sum of these expressions.
• The degree of a polynomial is the largest exponent of the variable.
• They are smooth and continuous, which means they have no sharp corners or jumps.

These functions play a significant role in calculus and algebra because they serve as a cornerstone for defining more complex functions. They tend to be easier to work with when calculating derivatives and solving equations. Understanding polynomial functions is essential for tasks such as finding critical numbers, evaluating function behavior, and performing integration.

Quadratic Equation Solving

Quadratic equations take the form \( ax^2 + bx + c = 0 \) and these can often be solved by factorization, completing the square, or using the quadratic formula. For the quadratic equation derived from the derivative \( 3x^2 + 12x - 15 = 0 \), we simplified it by dividing through by 3 to give \( x^2 + 4x - 5 = 0 \).

Solving by factoring involves rewriting the quadratic as a product of binomials.

• We look for two numbers that multiply to \(-5\) (the constant term) and add to \(4\) (the linear coefficient).
• The numbers \(5\) and \(-1\) fit this requirement, hence the factors are \((x + 5)(x - 1) = 0\).

The solutions to the equation are the values of \(x\) that make the expression zero, which are \( x = -5 \) and \( x = 1 \). Solving quadratic equations is essential for determining critical points where the derivative equals zero.

Factoring Technique

Factoring is a technique used to simplify algebraic expressions and solve equations. The goal is to rewrite the expression as a product of its factors. When solving for critical numbers of the derivative \( f'(x) = 3x^2 + 12x - 15 \), factoring allows us to simplify and resolve the quadratic expression \( x^2 + 4x - 5 = 0 \).

Here's how factoring works:

• First, identify two numbers that multiply to the constant term \(-5\) and add to the coefficient of \(x\), which is \(4\).
• The numbers are \(5\) and \(-1\), allowing us to write the quadratic as \((x + 5)(x - 1) = 0\).
• Using the Zero-Product Property, setting each factor equal to zero gives \(x = -5\) and \(x = 1\).

Factoring is a powerful algebraic technique that reveals the roots of quadratic equations, crucial for finding the critical points where polynomials change their behavior.