Problem 46 - University Calculus: Early Transcendentals

Question

Determine all critical points for each function. $$f(x)=6 x^{2}-x^{3}$$

Step-by-Step Solution

Answer

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

Step 1: Find the derivative

To find the critical points of the function, start by calculating the derivative of the function. The function given is \( f(x) = 6x^2 - x^3 \). Differentiate the function with respect to \( x \):\[ f'(x) = \frac{d}{dx}(6x^2 - x^3) = 12x - 3x^2 \]

Step 2: Set the derivative to zero

Critical points occur where the first derivative is zero or undefined. Since \( f'(x) = 12x - 3x^2 \), set this equal to zero to find the critical points:\[ 12x - 3x^2 = 0 \]

Step 3: Solve for x

Solve the equation from Step 2:Factor the expression:\[ 3x(4 - x) = 0 \]Set each factor to zero:1. \( 3x = 0 \) leads to \( x = 0 \)2. \( 4 - x = 0 \) leads to \( x = 4 \)Thus, the critical points are at \( x = 0 \) and \( x = 4 \).

Step 4: Verify the critical points

Verify the second derivative to determine the nature of these critical points. We will not calculate the second derivative explicitly here, but just confirm the points identified by the first derivative test.

Key Concepts

First Derivative

The first derivative of a function, often denoted as \( f'(x) \), provides us with important information about the function's rate of change. Calculating the first derivative is the initial step in identifying the critical points of the function. For the function \( f(x) = 6x^2 - x^3 \), the first derivative is calculated by differentiating each term separately:

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

This results in the final expression \( f'(x) = 12x - 3x^2 \). A function's critical points occur where its first derivative is zero or undefined. For our example, setting \( 12x - 3x^2 = 0 \), we solve for \( x \) to find these points. This process reveals where the function may have local maxima, minima, or saddle points.

Second Derivative Test

The second derivative test is a tool for determining the concavity of a function at its critical points. While our original step-by-step solution did not explicitly calculate it, the second derivative can indicate whether a critical point is a local maximum, minimum, or a point of inflection. The second derivative, \( f''(x) \), is found by differentiating \( f'(x) \) again:For \( f'(x) = 12x - 3x^2 \):

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

Thus, \( f''(x) = 12 - 6x \). By substituting the critical points into this second derivative:

• If \( f''(x) > 0 \), the function is concave upward, indicating a local minimum.
• If \( f''(x) < 0 \), the function is concave downward, suggesting a local maximum.

This test helps verify the nature of critical points beyond merely identifying them.

Polynomial Functions

Polynomial functions are fundamental in calculus and generally composed of terms of the form \( ax^n \), where \( n \) is a non-negative integer. For instance, \( f(x) = 6x^2 - x^3 \) is a polynomial of degree 3. Polynomial functions can be easily differentiated, as seen in calculating the first and second derivatives. The power rule is especially useful:

• To differentiate \( ax^n \), multiply by the power and subtract one from the exponent, resulting in \( anx^{n-1} \).

Polynomials are continuous and differentiable, which makes them straightforward when performing operations like integration and deriving the first or second derivatives. Their critical points and behavior are also fully determined by their derivatives, showcasing the importance of these concepts in understanding the function's graph.

Factorization in Calculus

Factorization is essential in calculus for simplifying expressions and solving equations, such as in finding critical points. For the derivative \( f'(x) = 12x - 3x^2 \), factorization was used to simplify and solve the equation \( 12x - 3x^2 = 0 \). Breaking this down:

• Factor out the greatest common factor, in this case \( 3x \), resulting in \( 3x(4 - x) = 0 \).
• Set each factor equal to zero, solving for \( x \).
• This yields the critical points \( x = 0 \) and \( x = 4 \).

Factorization reduces complexity and allows us to work with simpler forms of equations, making it easier to find solutions. It is a powerful technique for solving polynomial equations and interpreting critical points within calculus.