Problem 2 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

The function \(f(x)=x^{3}-3 x^{2}-24 x+5\) is an increasing function in the interval given below \(\quad\) IMP PET-1998] (a) \((-\infty,-2) \cup(4, \infty)\) (b) \((-2, \infty)\) (c) \((-2,4)\) (d) \((-\infty, 4)\)

Step-by-Step Solution

Answer

The function is increasing in \((- \infty, -2) 1 (4, \infty)\), so the answer is (a).

Step 1: Find the derivative

To determine where the function is increasing, we need to find the derivative of the function. The derivative of a function gives us the slope, and the function is increasing where this slope is positive.Given function: \( f(x) = x^3 - 3x^2 - 24x + 5 \).Derive the function with respect to \( x \):\[ f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(24x) + \frac{d}{dx}(5) \].This becomes:\[ f'(x) = 3x^2 - 6x - 24 \].

Step 2: Set the derivative to zero

To find points where the function changes from increasing to decreasing or vice versa, solve \( f'(x) = 0 \).Set the derivative equal to zero:\[ 3x^2 - 6x - 24 = 0 \].

Step 3: Solve the quadratic equation

Solve the quadratic equation \( 3x^2 - 6x - 24 = 0 \) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 3 \), \( b = -6 \), and \( c = -24 \).Substituting in these values:\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 3 \times (-24)}}{2 \times 3} \].This simplifies to:\[ x = \frac{6 \pm \sqrt{36 + 288}}{6} \].Further simplify to:\[ x = \frac{6 \pm \sqrt{324}}{6} \].Since \( \sqrt{324} = 18 \), the roots are:\[ x = \frac{6 + 18}{6} = 4 \] and \[ x = \frac{6 - 18}{6} = -2 \].

Step 4: Determine intervals of increase/decrease

Use test points to determine where the derivative is positive. Test intervals are based on the critical points obtained: \((-\infty, -2)\), \((-2, 4)\), and \((4, \infty)\).1. Choose a test point in each interval, like \( x = -3 \) for \((-\infty, -2)\), \( x = 0 \) for \((-2, 4)\), and \( x = 5 \) for \((4, \infty)\).2. Evaluate \( f'(x) \) at these points: - For \( x = -3 \), \( f'(-3) = 27 + 18 - 24 = 21 \) (positive). - For \( x = 0 \), \( f'(0) = -24 \) (negative). - For \( x = 5 \), \( f'(5) = 75 - 30 - 24 = 21 \) (positive).The function increases where \( f'(x) > 0 \). This happens in intervals \((-\infty, -2)\) and \((4, \infty)\).

Step 5: Select the correct answer

From Step 4, we concluded that the function is increasing where \( f'(x) > 0 \), i.e., in the intervals \((-\infty, -2)\) and \((4, \infty)\). Thus, the correct answer is option (a) \((-\infty, -2) 1 (4, \infty)\).

Key Concepts

Derivative

Derivatives are fundamental in calculus, providing us a powerful tool to analyze functions. Think of a derivative as a way to measure how a function changes. It tells us the **slope of the function** or how steep it is at any given point. When you see a graph, the derivative tells you how the graph climbs up or slides down.

For a function like the one given, which is a polynomial, finding the derivative involves applying basic rules like the power rule. The power rule states that if you have a term in the form of \( x^n \), its derivative is \( nx^{n-1} \).

Applying these rules to the function \( f(x) = x^3 - 3x^2 - 24x + 5 \), we get its derivative:

• \( f'(x) = 3x^2 - 6x - 24 \)

This new function, \( f'(x) \), tells us where the original function \( f(x) \) is increasing or decreasing. An important thing to remember is that wherever this derivative is positive, the original function is sloping upwards, or increasing.

Quadratic Equation

Quadratic equations are a type of polynomial equation of the form \( ax^2 + bx + c = 0 \). These equations often pop up when we set a derivative to zero to find critical points where the function changes direction.

In the exercise, we found the derivative and set it to zero:
\[ 3x^2 - 6x - 24 = 0 \]
To find the solutions or roots of this equation which help us identify key points on the graph of the function, we can use the **quadratic formula**:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• Here \( a = 3 \), \( b = -6 \), and \( c = -24 \).
• Plugging these into the formula gives the roots \( x = 4 \) and \( x = -2 \).

These roots are the points where the derivative switches signs, indicating a change in the increasing or decreasing nature of the function. Understanding quadratics and solving them is crucial for analyzing and interpreting these changes.

Function Increasing/Decreasing

For any function, understanding where it is increasing or decreasing provides great insights into its behavior and shape. Once we have the derivative, like \( f'(x) = 3x^2 - 6x - 24 \), we can use it to find these intervals of increase or decrease.

Here's how you can interpret these intervals:

• A function is **increasing** where its derivative is positive. This means the slope of the tangent is upward and the function value gets larger as you move along the x-axis.
• Conversely, a function is **decreasing** where its derivative is negative. This means the slope is downward, and the function value decreases.

To find these intervals in the exercise, test points were used:

• For \( x = -3 \), \( f'(-3) = 21 \) (positive), indicating an increasing interval.
• For \( x = 0 \), \( f'(0) = -24 \) (negative), indicating a decreasing interval.
• For \( x = 5 \), \( f'(5) = 21 \) (positive), indicating another increasing interval.

This method shows you where the function goes up or down and helps determine options like those given in the exercise. The function is increasing in the intervals \((-\infty,-2)\) and \((4, \infty)\), confirmed by positive derivative values. Understanding this concept helps in sketching graphs and predicting function behavior.