Problem 31 - Single Variable Calculus

Question

In Exercises \(25-40\), find the critical number \((s)\), if any, of the function. $$ h(x)=x^{4}-4 x^{3}+12 $$

Step-by-Step Solution

Answer

The critical numbers of the function \(h(x) = x^4 - 4x^3 + 12\) are \(x = 0\) and \(x = 3\).

Step 1: Find the derivative of the function

To find the critical numbers, we first need to find the derivative of the function \(h(x)\). The given function is \(h(x) = x^4 - 4x^3 + 12\). Using the power rule, for each term, we get the derivative as: \(h'(x) = 4x^3 - 12x^2\)

Step 2: Set the derivative to zero and find x

Now, we need to find the values of x for which \(h'(x) = 0\). Let's set the derivative to 0: \(4x^3 - 12x^2 = 0\) Simplify the equation by factoring out the common term: \((4x^2)(x - 3) = 0\)

Step 3: Solve for x

Now, we have a product of two factors equal to zero. By the zero product property, one of them must be zero. So, we have two cases: Case 1: \(4x^2 = 0\) Divide both sides by 4: \(x^2 = 0\) Therefore, \(x = 0\) (since \(0^2 = 0\)) Case 2: \(x - 3 = 0\) Add 3 to both sides: \(x = 3\)

Step 4: Write the solution

The critical numbers of the function \(h(x) = x^4 - 4x^3 + 12\) are \(x = 0\) and \(x = 3\).

Key Concepts

Derivative

The concept of a derivative is foundational in calculus. It helps us determine the rate at which a function is changing at any given point. To find a derivative, we need to apply rules of differentiation, and this crucially allows us to locate the critical numbers of a function. In this exercise, the function given is \( h(x) = x^4 - 4x^3 + 12 \). We found the derivative, \( h'(x) \), to be \( 4x^3 - 12x^2 \). This process allows us to move forward in identifying points where the function’s slope is zero, signaling potential critical numbers.

Finding the derivative is the first step towards understanding where a function may have peaks, troughs, or inflection points. By calculating the derivative, we are essentially finding a new function that describes the original function's rate of change at every point of its domain.

Power Rule

The power rule is a basic tool in calculus that simplifies the differentiation process. It is specifically handy for functions like polynomials, where each term consists of a variable raised to a power. The rule states that if you have a term \( x^n \), its derivative is \( nx^{n-1} \). Using this rule, we can efficiently find derivatives without the cumbersome process of incremental limits.

In the problem at hand, we applied the power rule to each term of \( h(x) = x^4 - 4x^3 + 12 \):

• For \( x^4 \), the derivative is \( 4x^3 \),
• For \( -4x^3 \), it is \( -12x^2 \),
• The derivative of constants, like 12, is zero.

This gives us the derivative \( h'(x) = 4x^3 - 12x^2 \). Understanding the power rule is crucial as it provides a quick way to compute derivatives, setting the stage for more complex calculations.

Factoring

Factoring is a method used to simplify expressions and solve equations, which is especially valuable when dealing with polynomials. In our exercise, after finding the derivative \( h'(x) = 4x^3 - 12x^2 \), we set it to zero to find critical numbers. To simplify solving \( 4x^3 - 12x^2 = 0 \), we factor the expression.

We look for a common factor in the terms, which in this case is \( 4x^2 \). By factoring it out, we get:

• \( 4x^2(x - 3) = 0 \)

Factoring transforms a complex polynomial equation into a simpler one, making it easier to apply the zero product property to find solutions. This step is essential in many algebraic manipulations, allowing us to solve for the variable of interest.

Zero Product Property

The zero product property is a fundamental concept in algebra that is used to solve equations set to zero. When a product of factors equals zero, at least one of the factors must be zero. This principle is critical for solving polynomial equations.

After factoring the derivative to \( 4x^2(x - 3) = 0 \), we apply the zero product property:

• In Case 1, \( 4x^2 = 0 \), solving gives \( x = 0 \).
• In Case 2, \( x - 3 = 0 \), solving gives \( x = 3 \).

Thus, the critical numbers of the function are \( x = 0 \) and \( x = 3 \). By using the zero product property, we can effectively determine the values of \( x \) that turn the product to zero, identifying points where the original function might have a change in direction or critical behavior.