Problem 71 - Calculus for JEE ADVANCED

Question

\(f^{\prime}(x)=-3 x^{2}-6 x-2<0 \forall x>0\).

Step-by-Step Solution

Answer

The derivative is negative for all \( x > 0 \).

Step 1: Understand the Problem

We are given a derivative of a function, \( f'(x) = -3x^2 - 6x - 2 \), and are asked to determine for which values of \( x \) the derivative is negative. Specifically, we need to evaluate \( f'(x) < 0 \) for \( x > 0 \).

Step 2: Set Up the Inequality

The expression \( f'(x) = -3x^2 - 6x - 2 \) is provided. We need to solve \( -3x^2 - 6x - 2 < 0 \).

Step 3: Solve the Quadratic Inequality

The inequality \( -3x^2 - 6x - 2 < 0 \) is a quadratic inequality. We first need to determine when \( f'(x) = 0 \). Calculate the roots of \( -3x^2 - 6x - 2 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = -3, b = -6, \) and \( c = -2 \).

Step 4: Calculate Discriminant

First, calculate the discriminant \( b^2 - 4ac \): \( (-6)^2 - 4(-3)(-2) = 36 - 24 = 12 \). This value is positive, indicating two real roots.

Step 5: Apply the Quadratic Formula

Using the quadratic formula, the roots are found by substituting \( a, b, \) and \( c \) into the formula: \( x = \frac{6 \pm \sqrt{12}}{-6} \). This can be simplified to: \( x = \frac{6 \pm 2\sqrt{3}}{-6} \), resulting in the roots \( x = -1 \pm \frac{\sqrt{3}}{3} \).

Step 6: Identify Intervals for x

The roots provide us with points: \( x_1 = -1 + \frac{\sqrt{3}}{3} \) and \( x_2 = -1 - \frac{\sqrt{3}}{3} \). However, we are only interested in \( x > 0 \). We determine where \( -3x^2 - 6x - 2 < 0 \) is true within the domain \( x > 0 \).

Step 7: Test Intervals and Define Solution

We can test points within the intervals defined by the roots to see where the inequality holds. Since we are only interested in \( x > 0 \), we should test values greater than \( 0 \). Given the parabola opens downwards (coefficient of \( x^2 \) is negative), \( -3x^2 - 6x - 2 < 0 \) for all \( x > 0 \). So the inequality is satisfied for all \( x > 0 \).

Key Concepts

Calculus Problems

In calculus, problems often revolve around understanding how a function behaves over certain intervals. One common type of problem involves determining the intervals where a derivative of a function is positive, negative, or zero. This is crucial because the sign of the derivative gives us insight into the increasing or decreasing behavior of a function over its domain.

When tackling calculus problems like these, it is important to carefully interpret the problem statements. For instance, when given a derivative like in this exercise, the task is to find where it is negative. This means we're looking for intervals where the function is decreasing.

Such problems often require us to work with inequalities and to have a firm grasp on solving them, as well as interpreting their solutions to reflect the behavior of the original function over its entire domain.

Quadratic Formula

The quadratic formula is a powerful tool used for finding the roots of quadratic equations of the form \( ax^2 + bx + c = 0 \). By solving for the roots, we can find the points where the parabola intersects the x-axis. These roots help us determine intervals for analyzing inequalities.

The quadratic formula is given by:

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

When using the formula, first calculate the discriminant, \( b^2 - 4ac \). The sign and value of the discriminant help determine the nature and number of the roots:

• If the discriminant is positive, the quadratic has two real and distinct roots.
• If it is zero, there is one real root (a double root).
• If negative, there are no real roots, only complex ones.

In the provided exercise, the discriminant was positive, leading to two real roots. These roots allowed us to breakdown the quadratic inequality into useful intervals, where solving these intervals shows where the derivative is negative.

Inequality Solutions

Solving quadratic inequalities involves finding the values of \(x\) that satisfy the inequality, crucial for understanding the behavior of functions. Once you find the roots using the quadratic formula, sketching a simple number line helps visualize solutions.

For example, if we determine roots are at \( x_1 \) and \( x_2 \), and the parabola opens downwards (as indicated by a negative \(a\)), the expression \( -3x^2 - 6x - 2 \) is less than zero between the two roots, assuming they were applicable in \(x > 0\).

To solve the inequality \( -3x^2 - 6x - 2 < 0 \), first factorize or use test points in intervals defined by the roots. Use test points to see where the inequality checks out as true.

Ultimately, inequality solutions can be intuitive with practice. Regularly test intervals, and if the parabola's direction matches the inequality, you've found your solution, as in the problem where all \( x > 0 \) satisfied the condition.