Problem 2 - A Textbook of Differential Calculus

Question

Let \(f: X \rightarrow Y, f(x)=\sin x+\cos x+2 \sqrt{2}\) is invertible, then \(X \rightarrow Y\) is/are (a) \(\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right] \rightarrow[\sqrt{2}, 3 \sqrt{2}]\) (b) \(\left[-\frac{\pi}{4}, \frac{3 \pi}{4}\right] \rightarrow[\sqrt{2}, 3 \sqrt{2}]\) (c) \(\left[-\frac{3 \pi}{4}, \frac{3 \pi}{4}\right] \rightarrow[\sqrt{2},-3 \sqrt{2}]\) (d) \(\left[-\frac{3 \pi}{4},-\frac{\pi}{4}\right] \rightarrow[\sqrt{2}, 3 \sqrt{2}]\)

Step-by-Step Solution

Answer

Option (a) \( \left[\frac{\pi}{4}, \frac{5 \pi}{4}\right] \rightarrow [\sqrt{2}, 3\sqrt{2}] \) is invertible.

Step 1: Determine the function range

The function given is \( f(x) = \sin x + \cos x + 2\sqrt{2} \). We recognize that the maximum value of \( \sin x + \cos x \) is \( \sqrt{2} \) and its minimum value is \( -\sqrt{2} \). Adding \( 2\sqrt{2} \) to these, the range of \( f(x) \) is \( [\sqrt{2}, 3\sqrt{2}] \).

Step 2: Determine invertibility criteria

For a function to be invertible, it must be one-to-one on the given domain. This means that within the specific interval for \( x \), the function must either be entirely increasing or decreasing, without repeating any output values.

Step 3: Analyze each option for monotonicity within domain

\( \sin x + \cos x \) is monotonic on the interval \([\frac{\pi}{4}, \frac{5\pi}{4}]\) since the derivative \( \cos x - \sin x \) shifts sign and maintains either increasing or decreasing behavior over this particular interval, ensuring invertibility on \([\frac{\pi}{4}, \frac{5\pi}{4}]\).

Step 4: Check specific intervals for invertibility

If we look at option (a), the interval \( \left[ \frac{\pi}{4}, \frac{5 \pi}{4} \right] \) covers one complete cycle where \( \sin x + \cos x \) is increasing, then decreasing, ensuring that it covers values \( [\sqrt{2}, 3\sqrt{2}] \), making the function invertible. Other options don't meet this strict monotonicity condition across their intervals based on \( \sin x + \cos x \)'s periodic behavior.

Key Concepts

Trigonometric Functions

Trigonometric functions like \(\sin x\) and \(\cos x\) are foundational in mathematics. They are periodic functions, meaning they repeat their values in regular intervals.
For both \(\sin x\) and \(\cos x\), this period is \(2\pi\). Every \(2\pi\) radians or 360 degrees, these functions will repeat their patterns.

• **Sine Function (\(\sin x\))**: It starts at 0, reaches its maximum of 1 at \(\frac{\pi}{2}\), back down to 0 at \(\pi\), decreases to \(-1\) at \(\frac{3\pi}{2}\), and back to 0 at \(2\pi\).
• **Cosine Function (\(\cos x\))**: It starts at 1, decreases to 0 at \(\frac{\pi}{2}\), goes to \(-1\) at \(\pi\), back up to 0 at \(\frac{3\pi}{2}\), and ends again at 1 at \(2\pi\).

Understanding how these functions behave between 0 and \(2\pi\) is crucial for analyzing their combinations, such as \(\sin x + \cos x\). In this exercise, \(\sin x + \cos x\) plays a key role, affecting the function's behavior and range determination.

Monotonicity

Monotonicity is an essential characteristic for determining whether a function is invertible. A function is monotonic if it is entirely non-increasing or non-decreasing over a certain interval.
This means:

• **Increasing Monotonicity**: If for any two points \(x_1\) and \(x_2\), where \(x_1 < x_2\), the inequality \(f(x_1) < f(x_2)\) holds true within that interval.
• **Decreasing Monotonicity**: If for any two points \(x_1\) and \(x_2\), where \(x_1 < x_2\), the inequality \(f(x_1) > f(x_2)\) holds true within that interval.

To determine invertibility, the interval must be one where the function \(\sin x + \cos x\) is monotonic. This is checked using its derivative \(\cos x - \sin x\). A change in the sign of this derivative over an interval indicates a change in monotonicity, suggesting the function may not be invertible. In our exercise, ensuring monotonicity between \([\frac{\pi}{4}, \frac{5\pi}{4}]\) verifies that the function can indeed be inverted on this interval.

Function Range

The range of a function is the set of all possible output values. Determining this range is crucial, especially when assessing invertibility and ensuring coverage of all potential outputs.
For the given function \(f(x) = \sin x + \cos x + 2\sqrt{2}\), we start by finding the range of \(\sin x + \cos x\).

• The maximum value for \(\sin x + \cos x\) is \(\sqrt{2}\), achieved when \(\sin x = \cos x\).
• The minimum value is \(-\sqrt{2}\), occurring when \(\sin x = -\cos x\).

By adjusting these values with an addition of \(2\sqrt{2}\), the range shifts to \([\sqrt{2}, 3\sqrt{2}]\). This interval signifies all achievable values from the given function due to the nature of \(\sin x\) and \(\cos x\)'s periodicity. Knowing this range helps identify which domain intervals maintain a one-to-one correspondence, essential for invertibility.