Problem 96 - Calculus and its applications

Question

Find the interval(s) for which \(f^{\prime}(x)\) is positive. $$ f(x)=\frac{1}{3} x^{3}-x^{2}-3 x+5 $$

Step-by-Step Solution

Answer

\(f^{\prime}(x)\) is positive on \((-\infty, -1)\) and \((3, \infty)\).

Step 1: Find the Derivative

To determine where the derivative is positive, we first need to find the derivative of the function. The function given is \(f(x) = \frac{1}{3}x^3 - x^2 - 3x + 5\). The derivative \(f^{\prime}(x)\) is found by differentiating each term of the function: \(f^{\prime}(x) = x^2 - 2x - 3\).

Step 2: Solve the Derivative for Zero

To find the intervals where \(f^{\prime}(x)\) is positive, we identify where \(f^{\prime}(x) = 0\). So, we solve \(x^2 - 2x - 3 = 0\) by factoring: \((x - 3)(x + 1) = 0\). Therefore, the critical points are \(x = 3\) and \(x = -1\).

Step 3: Test the Intervals

We need to test the sign of \(f^{\prime}(x)\) around the critical points \(x = -1\) and \(x = 3\). Consider the intervals \( (-\infty, -1) \), \( (-1, 3) \), and \( (3, \infty) \). Pick test points in each interval: \(x = -2\), \(x = 0\), and \(x = 4\).

Step 4: Evaluate the Derivative at Test Points

Evaluate \(f^{\prime}(x)\) at \(x = -2\): \((-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5\), which is positive. Evaluate \(f^{\prime}(x)\) at \(x = 0\): \((0)^2 - 2(0) - 3 = -3\), which is negative. Evaluate \(f^{\prime}(x)\) at \(x = 4\): \((4)^2 - 2(4) - 3 = 16 - 8 - 3 = 5\), which is positive.

Step 5: Determine Positive Intervals

From the evaluations, \(f^{\prime}(x) > 0\) on the intervals \((-\infty, -1)\) and \((3, \infty)\). Therefore, \(f^{\prime}(x)\) is positive for these intervals.

Key Concepts

Derivative

Understanding the derivative is crucial in the context of calculus, as it represents the rate at which a function's value changes. The derivative of a function provides insight into its slope at any given point.

When we talk about the derivative of a function like:\[ f(x) = \frac{1}{3}x^3 - x^2 - 3x + 5 \]we mean finding a function that describes how fast or slow this polynomial grows or declines.

To find the derivative, follow these steps:

• Differentiate each term separately.
• Apply the power rule: bring down the exponent as a multiplier and reduce the exponent by one. For example, for \( \frac{1}{3}x^3 \), the derivative is \( x^2 \).
• Combine all the differentiated terms for the complete derivative function.

Thus, the derivative for this function is:\[ f^{\prime}(x) = x^2 - 2x - 3 \]Understanding this derivative helps determine the behavior of \( f(x) \) over different intervals.

Critical Points

Critical points play a pivotal role in understanding where a function changes behavior, such as shifting from increasing to decreasing.

To find critical points, set the derivative equal to zero and solve for \(x\):\[ x^2 - 2x - 3 = 0 \]
These points are where the function's slope is zero, indicating a potential maximum, minimum, or neither. Solving this example:

• Factor the quadratic: \((x - 3)(x + 1) = 0\)
• Identify the critical points: \(x = 3\) and \(x = -1\)

At these points, \(f^{\prime}(x) = 0\), meaning the function may change direction.
Knowing critical points is essential for interval testing and understanding the overall behavior of the function.

Intervals

Determining intervals is all about where a function behaves in a particular way, such as increasing or decreasing.

Intervals are segments of the number line, formed around critical points, where the function's derivative is either positive or negative.
Let's break it down:

• Identify key intervals: from critical points, the function can be split into \((-\infty, -1)\), \((-1, 3)\), and \((3, \infty)\).
• Test a point within each interval to determine if the function's derivative is positive or negative.

For example, testing these critical intervals allows us to see where the function is increasing or decreasing.
This testing reveals:- In \((-\infty, -1)\) and \((3, \infty)\), the derivative is positive, indicating the function's growth.- In \((-1, 3)\), the derivative is negative, indicating a decrease.
Understanding intervals where the function increases or decreases allows deeper insights into the function's graph.

Quadratic Equation

Quadratic equations are a fundamental aspect of calculus, often appearing when dealing with polynomial derivatives set equal to zero.

The standard form of a quadratic equation is:\[ ax^2 + bx + c = 0 \]These equations can often be solved using factoring, completing the square, or the quadratic formula.

In our solution:

• The equation \(x^2 - 2x - 3 = 0\) reflects a quadratic form.
• We solved it by factoring: \((x - 3)(x + 1) = 0\).

Factoring separates the equation into two simpler expressions, enabling us to find the critical points.
Understanding how to handle quadratic equations is vital for identifying where functions like derivatives equal zero, which in turn helps to locate critical points and identify important function behaviors.