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Chapter 3: Problem 8

Find the critical numbers of each function. \(f(x)=x^{3}+6 x^{2}-36 x-60\)

Short Answer

Expert verified
The critical numbers are \(x = -6\) and \(x = 2\).

Step by step solution

01

Find the Derivative

To find the critical numbers of the function, we first need to find its derivative. Given the function \(f(x) = x^3 + 6x^2 - 36x - 60\), we differentiate it with respect to \(x\). Using the power rule, the derivative is: \(f'(x) = 3x^2 + 12x - 36\).
02

Set the Derivative Equal to Zero

Critical numbers occur where the derivative is zero or undefined. Here, we set \(f'(x) = 3x^2 + 12x - 36 = 0\). This is a quadratic equation that we will solve to find the values of \(x\).
03

Solve the Quadratic Equation

To solve the quadratic equation \(3x^2 + 12x - 36 = 0\), we can first divide every term by 3 to simplify: \(x^2 + 4x - 12 = 0\). We can then factor the quadratic: \((x + 6)(x - 2) = 0\). Thus, the solutions are \(x = -6\) and \(x = 2\).
04

Verify Solutions

The solutions \(x = -6\) and \(x = 2\) are where the derivative is zero, indicating potential critical numbers. Since the derivative is defined for all \(x\), these are the critical numbers for the function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
The concept of a derivative is central to calculus. It represents the rate at which a function is changing at any given point and provides information about the slope of the tangent line to the curve represented by the function.
To find a derivative, one typically uses differentiation rules, such as the power rule, product rule, quotient rule, or chain rule. In our case, to differentiate the function \(f(x) = x^3 + 6x^2 - 36x - 60\):
  • Apply the power rule: For any term in the form of \(ax^n\), the derivative is \(n \cdot ax^{n-1}\).
  • Use this rule on each term:
    • \(x^3\) becomes \(3x^2\).
    • \(6x^2\) becomes \(12x\).
    • \(-36x\) becomes \(-36\).
    • Constants like \(-60\) have a derivative of \(0\), since they don鈥檛 change.
Putting these together, we derive \(f'(x) = 3x^2 + 12x - 36\). Calculating derivatives helps in identifying critical points, locations where the first derivative is zero or undefined, highlighting changes in the function's behavior.
Quadratic Equation
Quadratic equations are polynomial equations of degree two. They take the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. In solving tasks like critical numbers, often you'll find a derivative leading to a quadratic equation.
Our equation \(3x^2 + 12x - 36 = 0\) is a classic example.
  • These equations can be solved using various methods:
    • Factoring, if the trinomial can be decomposed into a product of binomials.
    • Completing the square, a method of rearranging the equation into a perfect square trinomial.
    • The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), which provides solutions for any quadratic equation.
First, simplify wherever possible to make calculations easier. For \(3x^2 + 12x - 36 = 0\), we divide by 3 to simplify it to \(x^2 + 4x - 12 = 0\). Such simplification can make factoring or further methods easier to apply.
Factoring
Factoring is a mathematical process that involves writing an expression as a product of its constituent parts, called factors.
For quadratics like \(x^2 + 4x - 12 = 0\), factoring can be a powerful tool:
  • First, look for two numbers that multiply to the constant term (\(-12\)) and add to the linear coefficient (\(4\)).
  • In this example, the numbers \(6\) and \(-2\) fit these criteria.
  • This decomposition allows the expression to be written as \((x + 6)(x - 2)\).
Setting each factor equal to zero will provide the solutions:
  • \(x + 6 = 0\), which simplifies to \(x = -6\).
  • \(x - 2 = 0\), leading to \(x = 2\).
Factoring is not always straightforward, particularly if the factors are not immediately obvious, and more complex quadratics might require use of the quadratic formula. Nonetheless, it is a critical skill for simplifying and solving polynomial equations.

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