Problem 42 - Thomas Calculus

Question

In Exercises \(41-50,\) determine all critical points for each function. $$ f(x)=6 x^{2}-x^{3} $$

Step-by-Step Solution

Answer

The critical points are at \(x = 0\) and \(x = 4\).

Step 1: Find the derivative

To find critical points, we first need to take the derivative of the function. Start with the given function \( f(x) = 6x^2 - x^3 \). The derivative \( f'(x) \) is found using the power rule: 1. The derivative of \( 6x^2 \) is \( 12x \).2. The derivative of \( -x^3 \) is \( -3x^2 \).Thus, the derivative is \( f'(x) = 12x - 3x^2 \).

Step 2: Set the derivative to zero

To find critical points, set the derivative equal to zero and solve for \( x \):\[ 12x - 3x^2 = 0 \]

Step 3: Factor the equation

Factor the equation \( 12x - 3x^2 = 0 \) to simplify.\[ 3x(4 - x) = 0 \]This equation tells us that either \( 3x = 0 \) or \( 4 - x = 0 \).

Step 4: Solve for critical points

Solve each factor equation separately to find the critical points:1. From \( 3x = 0 \), we get \( x = 0 \).2. From \( 4 - x = 0 \), we get \( x = 4 \).Thus, the critical points are \( x = 0 \) and \( x = 4 \).

Key Concepts

derivative

Finding the derivative of a function is a fundamental concept in calculus. It tells us how a function changes as its input changes. The derivative gives the rate of change or the slope of the tangent line at any given point on the function. In the context of the problem, the derivative was used to locate the function's critical points, which are points where the slope is zero or undefined.

This means setting the derivative to zero allows us to find where the function's rate of change is neither increasing nor decreasing. It's a key step in identifying potential maxima or minima for the function, as these points frequently occur where the derivative equals zero.

power rule

The power rule is a basic technique in calculus used to differentiate functions of the form \( x^n \), where \( n \) is any real number. The rule states that the derivative of \( x^n \) is \( nx^{n-1} \). This simple rule allows us to quickly and easily find the derivative of polynomial functions without resorting to the formal definition of the derivative.

In the given exercise, for the term \( 6x^2 \), applying the power rule results in \( 12x \) by multiplying the exponent (2) by the coefficient (6) and then decreasing the exponent by one. Similarly, \( -x^3 \) becomes \( -3x^2 \) following the same steps. The power rule makes differentiating these terms straightforward and efficient.

factoring equations

Factoring is a crucial skill in algebra and calculus. It allows us to break down complex expressions into simpler ones that are easier to work with. In this exercise, after finding the derivative, the next step required setting it equal to zero. The resulting equation was \( 12x - 3x^2 = 0 \).

To solve for the critical points, we factor out the greatest common factor, which is \( 3x \) in this case. This simplification then gives us \( 3x(4 - x) = 0 \). Factoring simplifies the process because it breaks down the equation into smaller, readily solvable parts, and reveals potential solutions through each factor. It’s a bridge to the final step of solving for specific variables.

calculus

Calculus is the branch of mathematics focused on change. It involves concepts of derivatives and integrals. In this exercise, we used calculus to analyze the function \( f(x) = 6x^2 - x^3 \) by finding its critical points, which are crucial in determining where the function changes direction.

Critical points often correspond to the peaks and valleys of the function graph, known as maxima and minima. Calculus provides the tools to rigorously explore these changes by leveraging techniques like differentiation (to find critical points) and solving realistic problems by modeling dynamically changing situations.

solve for x

Solving for \( x \) is one of the primary goals when working with equations. It involves isolating \( x \) to understand where particular properties of a function occur, such as in the context of the critical points in this exercise.

Once the derivative \( 12x - 3x^2 = 0 \) was set to zero, factoring allowed us to express it as \( 3x(4 - x) = 0 \). The next step involved solving each factor equation separately: \( 3x = 0 \) gives \( x = 0 \), and \( 4 - x = 0 \) gives \( x = 4 \). These solutions indicate where the function \( f(x) = 6x^2 - x^3 \) has its critical points, illustrating what solving for \( x \) typically accomplishes in calculus.