Problem 66 - Calculus Early Transcendentals

Question

Find the critical points of \(f .\) Assume a is a constant. $$f(x)=x^{3}-3 a x^{2}+3 a^{2} x-a^{3}$$

Step-by-Step Solution

Answer

Answer: The critical point of the function is \(x = a\).

Step 1: Find the derivative of the function

We need to find the derivative of the function with respect to \(x\): $$f'(x) = \frac{d}{dx}(x^3 - 3ax^2 + 3a^2x - a^3)$$ Using the power rule, we then get: $$f'(x) = 3x^2 - 6ax + 3a^2$$

Step 2: Set the derivative to zero and solve for x

To find the critical points of the function, we need to set the derivative equal to zero and then solve for \(x\): $$0 = 3x^2 - 6ax + 3a^2$$ You can notice that 3 is a common factor, divide the expression by 3: $$0 = x^2 - 2ax + a^2$$ Now the expression looks like a quadratic equation \(ax^2 + bx + c = 0\). To solve for \(x\), we can use the quadratic formula: $$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ Substitute \(a = 1\), \(b=-2a\), and c = \(a^2\): $$x= \frac{-(-2a) \pm \sqrt{(-2a)^2-4(1)(a^2)}}{2(1)}$$

Step 3: Simplify the expression

Simplify the expression by plugging in the values of \(a\), \(b\), and \(c\): $$x= \frac{2a \pm \sqrt{4a^2-4a^2}}{2}$$ Notice that the discriminant, \(\sqrt{4a^2-4a^2}\), is equal to 0. This means that there is only one unique solution to the quadratic equation: $$x=\frac{2a}{2}$$ Simplify to get: $$x = a$$

Step 4: State the critical point

The critical point of the function for any value of the constant \(a\) is: $$x=a$$

Key Concepts

Derivative

Derivatives play a crucial role in calculus, particularly when identifying critical points of functions. A derivative is a mathematical tool used to calculate the rate at which a function's value changes with respect to changes in another variable. In simpler terms, it tells us how a function is changing at any given point. For our particular function \( f(x) = x^3 - 3ax^2 + 3a^2x - a^3 \), we start by taking the derivative with respect to \( x \).

To find the derivative, we apply the

• "Power Rule," which provides an easy way to differentiate functions of the form \( x^n \), resulting in \( n \cdot x^{n-1} \).

Applying this rule to the terms yields the derivative \( f'(x) = 3x^2 - 6ax + 3a^2 \). The values obtained from the derivative allow us to determine how the original function is behaving at any particular \( x \)-value. Specifically, it helps us find the points where the function changes its increasing or decreasing nature, known as critical points.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in a single variable \( x \) with the form \( ax^2 + bx + c = 0 \). In our problem, the derivative \( f'(x) = 3x^2 - 6ax + 3a^2 \) is simplified by factoring out the 3, resulting in \( x^2 - 2ax + a^2 = 0 \).

To find the critical points, we must solve this quadratic equation. Recognizing the standard form, we can apply the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

where \( a = 1 \), \( b = -2a \), and \( c = a^2 \) per our simplified equation. Through substitution, we find \( x = a \). Since the discriminant (\( b^2 - 4ac \)) equals zero, there is one unique solution, indicating a double root at this critical point.

Power Rule

The Power Rule is a fundamental principle in calculus used to find derivatives of power functions. It simplifies the differentiation process by stating that if \( f(x) = x^n \), then the derivative \( f'(x) \) is \( n \cdot x^{n-1} \).

When we apply the Power Rule to the function \( f(x) = x^3 - 3ax^2 + 3a^2x - a^3 \), it allows us to handle each term independently:

• For \( x^3 \), the derivative is \( 3x^2 \).
• For \(-3ax^2 \), applying the rule gives \(-6ax\).
• The term \(3a^2x\) results in a derivative of \(3a^2\).
• Lastly, the constant term \(-a^3\) simply vanishes, as constants have a derivative of zero.

These calculations culminate in the derivative \( f'(x) = 3x^2 - 6ax + 3a^2 \), illustrating the power of this rule in simplifying functions to identify their behavior efficiently.