Problem 299 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find the values of \(x\) for which the function \(f(x)=1+2 \sin x+3 \cos ^{2} x, \quad 0 \leq x \leq \frac{2 \pi}{3}\) has maxima or minima. Also find the values of the function at these extremum. \\{Ans. minima at \(x=\frac{\pi}{2}, f\left(\frac{\pi}{2}\right)=3\), maxima at \(\left.x=\sin ^{-1} \frac{1}{3}, f\left(\sin ^{-1} \frac{1}{3}\right)=\frac{13}{3}\right\\}\)

Step-by-Step Solution

Answer

The function \(f(x) = 1+2 \sin x+3 \cos ^{2} x\) reaches its minimum at \(x=\pi/2\) and \(f(x) = 3\), and its maximum at \(x=\sin^{-1}(1/3)\) and \(f(x) = 13/3\).

Step 1: Find the derivative

Find the derivative of the function \(f(x) = 1+2 \sin(x) +3 \cos ^{2}(x)\). The derivative of \(f(x)\) is \(f'(x) = 2 \cos(x) - 6 \cos(x)\sin(x)\).

Step 2: Find critical points

Set the derivative equal to zero and solve for \(x\). This yields two solutions, \(x = \pi/2\) and \(x = \sin^{-1}(1/3)\).

Step 3: Identify the extremum points

To determine whether each critical point is a maximum or a minimum, use the second derivative test. The second derivative of \(f(x)\) is \(f''(x) = -2 \sin(x) - 6\cos(x)^2 + 6\sin(x)\). Plug each critical point into \(f''(x)\). A negative result indicates a local maximum, while a positive result indicates a local minimum. At \(x = \pi/2\), \(f''(x) = -2\) - a negative number which means it's a local maximum. At \(x = \sin^{-1}(1/3)\), \(f''(x) = 2\sqrt{8/9}\), a positive number which means it is a local minimum.

Step 4: Find the values of the function at extrema

Substitute the extremum points into the original function to find their corresponding function values. At \(x = \pi/2\), \(f(x) = 3\). At \(x = \sin^{-1}(1/3)\), \(f(x) = 13/3\).

Key Concepts

Critical Points

Critical points in the landscape of trigonometric functions are the bread crumbs that lead us to the local extremum (peaks and troughs) of a function. Imagine you're on a hillside trying to find the highest or lowest points; critical points are where the slope of the function flattens, indicating potential maxima or minima.

In mathematical terms, to locate these points in a function like our exercise's function, we find where the derivative of the function equals zero or does not exist. In our case, setting the derivative (which is derived by applying the rules of differentiation to each trigonometric part of the function) equal to zero provides the places where the function's slope is zero. These points provide the first hint whether we have a hilltop or a valley floor - a maximum or a minimum.

Second Derivative Test

The second derivative test is like the detective's magnifying glass, examining the curvature of a function at critical points to determine if they are maxima or minima. Think of it as checking whether the path ahead bends upwards or downwards around those flat spots.

Applying the Second Derivative Test

For our function, after finding the critical points, we take the second derivative. Then, we substitute the critical points into it. If the outcome is positive, the curvature is upward, revealing a local minimum. Conversely, if the result is negative, the curvature arches downward, uncovering a local maximum. For instance, when we examine the function at the critical point \( x = \frac{\pi}{2} \) using the second derivative, we get a negative sign, signifying a maximum at this point.

Maxima and Minima

Maxima and minima, the high and low notes of a function's melody, represent the apex of a hill and the bottom of a valley, respectively, in practical scenarios. They play a crucial role in calculus and real-world applications like finding optimum quantities for maximum profit or minimum cost.

In our exercise, we use the critical points and the second derivative test to identify these intricate points of the function: a minimum at \( x = \sin^{-1}(1/3) \) where the function's value is \( \frac{13}{3} \), and a maximum at \( x = \frac{\pi}{2} \) where the function's value is 3. It's essential to plug critical points back into the original function to find these specific function values, ensuring we have the full picture of the function's behavior at these key points.