Question

Find the domain and range of the function. $$ f(t)=\sec \frac{\pi t}{4} $$

Short answer

The domain of the function \(f(t)=\sec \frac{\pi t}{4}\) is all real numbers 't' except those satisfying \(t = 4n + 2\) for any integer n. The range of the function is \(-\infty \leq y < -1\) union \(1 \leq y < \infty\).

Step-by-step solution

Identify the Original Function

Identify that the original function is \(f(t)= \sec \frac{\pi t}{4}\) which is the secant of \(\frac{\pi t}{4}\).

Find the Domain

The domain is all real numbers 't' such that \(\frac{\pi t}{4}\) is not equal to \((2n + 1)\frac{\pi}{2}\) for any integer n. This means we must exclude values of 't' for which cos(\(\frac{\pi t}{4}\)) = 0. Solving the equation \(\frac{\pi t}{4}\) = \((2n + 1)\frac{\pi}{2}\), we get t = \(4n + 2\). Therefore, the domain of the function is \(t \neq 4n + 2\) for every integer n.

Find the Range

The range of the secant function is \(-\infty \leq y < -1\) and \(1 \leq y \leq \infty\). This is because sec(x) is 1/cos(x) and cos(x) ranges between -1 and 1 and hence, their reciprocal will range between -∞ and -1 and 1 to ∞.

Key concepts

Domain and Range of a Function

Understanding the domain and range of a function is like identifying the playground where the function can operate. The domain refers to all the possible input values (typically 'x' or in this case, 't') a function can accept without causing mathematical complications, like division by zero or taking a square root of a negative number in real number systems.

For our secant function, we want to make sure we're not dividing by zero since secant is defined as the reciprocal of the cosine function. So, we eliminate any 't' value that makes \( \cos\frac{\pi t}{4} \) equal zero, resulting in a domain that excludes the values \( 4n + 2 \) for any integer 'n'.

On the flip side, the range represents the set of possible output values our function can produce. Since secant is the inverse of cosine, and cosine values are between -1 and 1, the secant function will have a range of \( -\infty \leq y < -1 \) and \( 1 \leq y \leq \infty \). Just think: you cannot have a reciprocal of zero, which translates to these function values jumping from negative to positive infinity at the points where the cosine function would have crossed zero.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are cornerstones of calculus, especially when dealing with periodic phenomena or circular motion. The secant function, represented by \( \sec(x) \), is one of these trigonometric functions. It's essentially the reciprocal of the cosine function, \( \sec(x) = \frac{1}{\cos(x)} \).

Likewise, for \( f(t) = \sec \frac{\pi t}{4} \), we are looking at a secant function with a period that has been adjusted by the coefficient \( \frac{\pi}{4} \). Understanding how to graph these functions and their transformations is critical when solving calculus problems involving trigonometric identities, integrals or derivatives. For example, knowing that secant has vertical asymptotes where cosine hits zero can help anticipate the function's behavior, which is essential for accurate graphing and analysis.

Calculus Problem Solving

Calculus problem solving often involves a step-by-step approach to break down complex problems into manageable parts. When faced with functions, especially trigonometric ones, it's vital to first identify its properties: its domain, range, peaks, valleys, asymptotes, and periods. After grasping these characteristics, calculus tasks such as differentiation and integration become more intuitive.

For instance, solving for the domain and range sets the stage, and we proceed by analyzing the function's behavior, calculating limits, derivatives, or integrals as needed. It's like putting together pieces of a puzzle - each element needs to be in the right place. By combining analytical skills with an understanding of trigonometry and calculus fundamentals, problems that initially seem daunting are broken down and solved with precision.