Problem 36 - Calculus Early Transcendentals

Question

Show that \(f\) is continuous on \((-\infty, \infty)\) \(f(x)=\left\\{\begin{array}{ll}{\sin x} & {\text { if } x<\pi / 4} \\ {\cos x} & {\text { if } x \geqslant \pi / 4}\end{array}\right.\)

Step-by-Step Solution

Answer

The function \( f(x) \) is continuous for all real \( x \).

Step 1: Check Continuity for \( x < rac{\pi}{4} \)

For the interval \( x < \frac{\pi}{4} \), the function is defined as \( f(x) = \sin x \). The sine function, \( \sin x \), is continuous for all real numbers \( x \). Thus, \( f \) is continuous on the interval \((-\infty, \frac{\pi}{4})\).

Step 2: Check Continuity for \( x \geq \frac{\pi}{4} \)

For the interval \( x \geq \frac{\pi}{4} \), the function is defined as \( f(x) = \cos x \). The cosine function, \( \cos x \), is also continuous for all real numbers \( x \). Thus, \( f \) is continuous on the interval \((\frac{\pi}{4}, \infty)\).

Step 3: Check Continuity at \( x = \frac{\pi}{4} \)

To establish continuity at \( x = \frac{\pi}{4} \), check the limit from both sides and ensure they are equal to the function value at that point. Calculate:1. \( \lim_{x \to \pi/4^-} f(x) = \lim_{x \to \pi/4^-} \sin x = \sin(\pi/4) = \frac{\sqrt{2}}{2} \).2. \( \lim_{x \to \pi/4^+} f(x) = \lim_{x \to \pi/4^+} \cos x = \cos(\pi/4) = \frac{\sqrt{2}}{2} \).Both limits are equal: \( \frac{\sqrt{2}}{2} \).3. The function value at \( x = \pi/4 \) is \( f(\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2} \).Since both the left-hand and right-hand limits and the function value are equal, \( f \) is continuous at \( x = \pi/4 \).

Step 4: Conclusion

Since \( f \) is continuous in each defined interval and remains continuous at \( x=\frac{\pi}{4} \) where the function expression changes, \( f \) is continuous over \((-ity, ity)\).

Key Concepts

Continuous Functions

Continuous functions are those that you can draw without lifting your pencil from the paper. If a function is continuous at a point, then it has no breaks, holes, or jumps at that point.

• A function is continuous over an interval if it is continuous at every point within that interval.
• Mathematically, a function \( f(x) \) is continuous at a point \( x = a \) if the following conditions are satisfied:
• The function is defined at \( x = a \), meaning \( f(a) \) exists.
• The left-hand limit at \( x = a \) equals the right-hand limit.
• The value of the function at that point is equal to the limit: \( \lim_{{x \to a}} f(x) = f(a) \).

In the given exercise, the function \( f(x) \) is composed of \( \sin x \) and \( \cos x \). Both sine and cosine are continuous over the entire set of real numbers. So, ensuring continuity for \( f(x) \) at internal intervals is straightforward. The only challenge is ensuring continuity at the transitional point \( x = \frac{\pi}{4} \).

Limits

Limits are essential for determining the behavior of a function as its input approaches a particular point. They tell us what value a function's output approaches as the input points get closer to a specific point.

• To ascertain continuity, limits help by indicating if the approach from the left (\( x \to a^- \)) and the right (\( x \to a^+ \)) lead to the same function value.
• If both the left-hand limit and the right-hand limit at a point are the same and equal to the function value at that point, the function is continuous there.

In this exercise, we used limits to confirm continuity at \( x = \frac{\pi}{4} \). We evaluated \( \lim_{{x \to \frac{\pi}{4}^-}} \sin x \) and \( \lim_{{x \to \frac{\pi}{4}^+}} \cos x \) and found them both equal to \( \frac{\sqrt{2}}{2} \). This confirmed that the function \( f(x) \) is continuous at that point.

Composite Functions

When we encounter functions defined by different sub-functions across various intervals, such as in this exercise, these are called composite functions. Here the function \( f(x) \) reads as a composite of \( \sin x \) and \( \cos x \) depending on the interval you're evaluating.

• At any point, the chosen segment of the composite function determines its output based on the condition it meets.
• To ensure continuity in composite functions, verify the continuity in individual segments and specifically at boundary points where the piecewise rules change.

In the given function, \( f(x) = \sin x \) for \( x < \frac{\pi}{4} \) and \( f(x) = \cos x \) for \( x \geq \frac{\pi}{4} \). Checking continuity involves confirming these segment behaviors and their transition at \( x = \frac{\pi}{4} \). This thorough check guarantees that the entire function remains continuous over \(( -\infty, \infty)\).