Problem 52 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(y=x^{3}-3 x^{2}\)

Step-by-Step Solution

Answer

The critical points are (0, 0) and (2, -4).

Step 1: Find the Derivative

To find critical points, we first need to find the derivative of the function. The given function is \( y = x^3 - 3x^2 \). The derivative of \( y \) with respect to \( x \) is \( y' = \frac{d}{dx}(x^3 - 3x^2) = 3x^2 - 6x \).

Step 2: Set the Derivative to Zero

To find the critical points, set the derivative equal to zero: \( 3x^2 - 6x = 0 \).

Step 3: Solve the Derivative Equation

Factor the equation \( 3x^2 - 6x = 0 \) as follows: \( 3x(x - 2) = 0 \). This gives the solutions \( x = 0 \) and \( x = 2 \). These are the critical points.

Step 4: Find Function Values at Critical Points

Substitute the critical points back into the original equation to find the y-values: \( y(0) = 0^3 - 3(0)^2 = 0 \) and \( y(2) = 2^3 - 3(2)^2 = 8 - 12 = -4 \).

Step 5: Conclusion on Critical Points

The critical points of the function \( y = x^3 - 3x^2 \) are \( (0, 0) \) and \( (2, -4) \). These points indicate where the slope of the tangent is zero, which could be a minimum or maximum or a point of inflection.

Key Concepts

Understanding Derivatives

Derivatives are a core concept in calculus that describe how a function changes as its input changes. Essentially, they represent the rate of change or the slope of a function at any given point. They are crucial for analyzing functions because they help identify where a function increases or decreases. In our exercise, we calculated the derivative of the function \( y = x^3 - 3x^2 \). This derivative is \( y' = 3x^2 - 6x \). This tells us the slope of the function at any point \( x \).

• Using the power rule, \( f(x) = ax^n \) becomes \( f'(x) = anx^{n-1} \).
• This is how \( x^3 \) becomes \( 3x^2 \) and \(-3x^2 \) becomes \(-6x \).

Knowing how to calculate derivatives allows us to explore and anticipate behavior in functions, especially when looking for critical points, which are points where the function changes direction.

Identifying Critical Points

Critical points are significant in calculus because they reveal where a function's slope is zero or undefined. In simpler terms, they show where a function might stop increasing and start decreasing, or vice versa. The critical points correspond to the maximums, minimums, or points of inflection of a function.To find these points, you set the derivative equal to zero. This is because a zero derivative indicates a horizontal slope, which is crucial for determining changes in direction:

• For our function, \( 3x(x - 2) = 0 \) leads to critical points at \( x = 0 \) and \( x = 2 \).
• By substituting these values back into the original function, we get \( y(0) = 0 \) and \( y(2) = -4 \).

These critical points \((0, 0)\) and \((2, -4)\) are candidates for extreme values or possible changes in the function's trend.

Conducting Function Analysis

Function analysis involves examining the critical points and the overall behavior of a function to identify its key features. Once we have critical points, like in our exercise, the next step is often to determine if these points are peaks, troughs, or neither.Function analysis provides valuable insights:

• By the continuity of the derivative, observe intervals where the derivative changes sign to identify local maxima or minima.
• For \( y = x^3 - 3x^2 \), you can use the second derivative test to further classify these points.
• If the second derivative at a critical point is positive, it suggests a local minimum; if it’s negative, a local maximum.

In our example, after finding \((0,0)\) and \((2,-4)\), function analysis aids in establishing whether these are the lowest or highest points, ensuring better understanding of the function’s dynamics and flow.