Problem 19 - Calculus

Question

Graph \(y=x^{3}+x^{2}-x\) from \(x=-2\) to \(x=2\) and estimate where it is decreasing. Check the transition points by solving \(d y / d x=0\).

Step-by-Step Solution

Answer

The function is decreasing for \( x \) in \((-1, \frac{1}{3})\).

Step 1: Find the Derivative

To find where the function is decreasing, we need its first derivative. Given the function \( y = x^3 + x^2 - x \), differentiate with respect to \( x \) to get:\[ \frac{dy}{dx} = 3x^2 + 2x - 1 \]

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find critical points:\[ 3x^2 + 2x - 1 = 0 \]Solve this quadratic equation using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 3 \), \( b = 2 \), and \( c = -1 \). This gives:\[ x = \frac{-2 \pm \sqrt{4 + 12}}{6} \]\[ x = \frac{-2 \pm 4}{6} \]The solutions are \( x = \frac{1}{3} \) and \( x = -1 \).

Step 3: Investigate the Intervals

Determine the sign of \( \frac{dy}{dx} \) in the intervals determined by the critical points to identify where \( y = x^3 + x^2 - x \) is decreasing.- Test the interval \((-\infty, -1)\): Pick \( x = -2 \), \( 3(-2)^2 + 2(-2) - 1 = 12 - 4 - 1 = 7 \) (positive)- Test the interval \((-1, \frac{1}{3})\): Pick \( x = 0 \), \( 3(0)^2 + 2(0) - 1 = -1 \) (negative)- Test the interval \((\frac{1}{3}, \infty)\): Pick \( x = 1 \), \( 3(1)^2 + 2(1) - 1 = 3 + 2 - 1 = 4 \) (positive)This shows the function decreases on \((-1, \frac{1}{3})\).

Step 4: Graph the Function

To visually confirm, graph \( y = x^3 + x^2 - x \) in the range \( x = -2 \) to \( x = 2 \) and note the decreasing interval \( x \in (-1, \frac{1}{3}) \). Look for parts of the graph where the slope transitions from positive to negative or negative to positive.

Key Concepts

Understanding the Derivative

The derivative of a function is a concept from calculus that represents the rate at which a function's output changes with respect to changes in its input. It's a fundamental tool in calculus, often thought of as a way to calculate the slope of the tangent line to the curve of a function at a given point.
In simpler terms, the derivative tells us how steep the graph of the function is at any point.
An important application of the derivative is finding when a function is increasing or decreasing.

• If the derivative is positive, the function is increasing at that point.
• If the derivative is negative, the function is decreasing at that point.

To find the derivative, we apply differentiation rules. For the function given, which is polynomial, we differentiate term by term. This approach is straightforward because each term is of the form of a power of x.

Finding Critical Points

Critical points of a function are those points where the derivative is zero or undefined. These points are important as they can indicate potential maxima, minima, or inflection points of the function.
To find the critical points of a function, follow these steps:

• Take the derivative of the function.
• Set the derivative equal to zero and solve for the variable.

In the worked example, the derivative \( \frac{dy}{dx} = 3x^2 + 2x - 1 \) is a quadratic equation. Setting it to zero yields \( 3x^2 + 2x - 1 = 0 \), which we solve using algebraic methods like factoring or the quadratic formula to find the critical points. These tell us where the slope transitions and give insights into the behavior of the function.

Applying the Quadratic Formula

The quadratic formula is an essential mathematical formula used to solve quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The formula provides the solutions for \( x \):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In this expression, \( a \), \( b \), and \( c \) are coefficients of the quadratic equation. The formula can find the roots of any quadratic equation, whether it can be factored easily or not.

• It calculates the number of real roots based on the discriminant \( b^2 - 4ac \).
• If the discriminant is positive, there are two distinct real roots.
• If zero, there's exactly one real root.
• If negative, the roots are complex.

In the problem context, substituting \( a = 3 \), \( b = 2 \), and \( c = -1 \) into the formula yields the critical points at \( x = \frac{1}{3} \) and \( x = -1 \), critical in analyzing the function's behavior.

The Practice of Function Graphing

Graphing functions is a powerful way to visualize and understand their behavior. It provides insight into the properties of the function, such as where it increases, decreases, and where the critical points are located.
To graph a function, you typically:

• Determine its critical points and any intercepts with the axes.
• Analyze intervals between critical points to check where the function increases or decreases.
• Sketch the curve, connecting these intervals smoothly.

For the function \( y = x^3 + x^2 - x \), plotting between \( x = -2 \) and \( x = 2 \) shows visually the intervals of increase and decrease. In this example, graphing helps to confirm the analytical results—where the function decreases in the interval \((-1, \frac{1}{3})\), indicated by a downslope on the graph.