Problem 57 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Does the function \(f(x)=\frac{x^{3}}{4}-\sin \pi x+3\) take the value \(2 \frac{1}{3}\) within the interval \([-2,2]\) ?

Step-by-Step Solution

Answer

Yes, the function \(f(x) = \frac{x^3}{4} - \sin \pi x + 3\) takes the value \(2\frac{1}{3}\) within the interval \([-2, 2]\).

Step 1: Set the function equal to the given value

We have to check if the function \(f(x) = \frac{x^3}{4} - \sin \pi x + 3\) takes the value \(2\frac{1}{3}\) within the interval \([-2, 2]\). So let's set the function equal to the given value and solve for x: \(\frac{x^3}{4} - \sin \pi x + 3 = 2\frac{1}{3}\)

Step 2: Simplify the equation

To simplify the equation, we can first convert \(2\frac{1}{3}\) into an improper fraction, in this case, \(7/3\): \(\frac{x^3}{4} - \sin \pi x + 3 = \frac{7}{3}\) Now let's subtract 3 from both sides of the equation: \(\frac{x^3}{4} - \sin \pi x = -\frac{2}{3}\)

Step 3: Graph the function and find its intersection with the given value

Now that we have the simplified equation, we can graph the function \(f(x) = \frac{x^3}{4} - \sin \pi x\), and also graph the horizontal line for value \(-\frac{2}{3}\), and then visually determine if the function intersects with the horizontal line within the given interval \([-2, 2]\). If it does, that means there is a value of x in the given interval for which the function takes the value \(2\frac{1}{3}\).

Step 4: Determine whether the function intersects with the given value within the interval

Upon graphing the function and the horizontal line, we can see that the function \(f(x) = \frac{x^3}{4} - \sin \pi x\) intersects with the horizontal line for \(-\frac{2}{3}\) within the interval \([-2, 2]\). Therefore, the function \(f(x) = \frac{x^3}{4} - \sin \pi x + 3\) does take the value \(2\frac{1}{3}\) within the interval \([-2, 2]\).

Key Concepts

Graphical Method in Calculus

The graphical method is a visual approach frequently used in calculus to understand and solve problems concerning functions. Through this method, one can plot the function on a coordinate system and examine its properties, such as continuity, domain, range, and points of intersection.
In our exercise, we use the graphical method to determine if the function \(f(x) = \frac{x^3}{4} - \sin \pi x + 3\) reaches a specific value within an interval. By plotting the function \(f(x)\) and drawing a horizontal line at \(-\frac{2}{3}\) which represents the simplified target value \(2\frac{1}{3}\), we visualize where the function's curve intersects this line within the interval \([-2, 2]\).
This intersection indicates where \(f(x)\) achieves the given value. Using applications like graphing calculators or computer software, this method provides a quick and effective way to spot the solution without the need for complex algebraic manipulations.

Solving Algebraic Equations

Algebraic equations form the backbone of many mathematical concepts and are particularly essential in calculus. To solve the algebraic equations effectively, one generally isolates the variable, simplifies the terms, and performs arithmetic operations until the solution is found. However, when these equations include trigonometric and polynomial expressions, as in our exercise, the task becomes more challenging.
Taking the initial function value equation, \(\frac{x^3}{4} - \sin \pi x + 3 = 2\frac{1}{3}\), and simplifying it to \(\frac{x^3}{4} - \sin \pi x = -\frac{2}{3}\), we prepare for the graphical method. However, traditionally, this equation would require one to isolate the variable \(x\) and solve for it. In cases where the equation can't be simplified easily, numerical methods, such as the Newton-Raphson method or interval bisection, could be employed to approximate the solution.

Trigonometric Functions in Calculus

Trigonometric functions play a central role in calculus, providing a means to model periodic phenomena and contributing to various forms of function analysis. In the context of our exercise, the \(\sin \pi x\) term gives the function a periodic characteristic, complicating the solution process.
Understanding the properties of trigonometric functions—such as their periods, amplitudes, and phase shifts—is crucial when working with them in calculus problems. These properties allow us to predict the behaviour of the function across different intervals. For the function \(f(x) = \frac{x^3}{4} - \sin \pi x + 3\), we need to ascertain its values over \([-2, 2]\), requiring comprehension of how the cubic and trigonometric parts interact over the given range. Solving and analyzing such functions often involve graphical displays to aid in understanding and to validate algebraic solutions.