Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

State whether the indicated function is continuous at \(3 .\) If it is not continuous, tell why. $$ g(t)=\sqrt{t-4} $$

Step-by-Step Solution

Answer

The function is not continuous at 3 because 3 is not in its domain.

Step 1: Evaluate the Function's Domain

The function \( g(t) = \sqrt{t - 4} \) is defined only when under the square root is non-negative. Therefore, \( t - 4 \geq 0 \), which implies \( t \geq 4 \). Hence, the domain of \( g(t) \) is \([4, \infty)\).

Step 2: Check if 3 is within the Domain

Since 3 is less than 4, it does not belong to the domain \([4, \infty)\) of the function \( g(t) \).

Step 3: Determine Continuity at t=3

For a function to be continuous at a point, that point must be within the function's domain. Since \( t = 3 \) is not in the domain of \( g(t) \), the function is not continuous at \( t = 3 \).

Key Concepts

Function Domain

In calculus, understanding a function's domain is crucial when we talk about continuity. The domain of a function refers to all the possible input values (often represented by 'x' or 't') for which the function is defined without any restrictions. For the function \( g(t) = \sqrt{t - 4} \), we need the expression inside the square root to be non-negative because the square root of a negative number is not real in typical real-valued function analysis.

This means we solve the inequality \( t-4 \geq 0 \), simplifying to \( t \geq 4 \). As a result, the domain of \( g(t) \) is \([4, \infty)\). This tells us that the function is defined, and thus can only accept inputs (or t-values) that are equal to or greater than 4. Understanding a function's domain helps determine where the function is valid and continuous.

Square Root Function

Square root functions have their own quirks or properties that affect their study in calculus. A square root function is of the form \( f(x) = \sqrt{x} \) or \( g(t) = \sqrt{t-4} \) in this case. The main characteristic of the square root function is that it produces real and non-negative outputs, but only when its input is also non-negative.

When analyzing \( g(t) = \sqrt{t - 4} \), we must ensure that the expression \( t - 4 \) inside the square root is not negative, which is why t must be at least 4. The square root function starts at the point \( t = 4 \) and continues to increase indefinitely. The understanding of this helps identify when and why the function might become undefined, hence showing why \( g(t) \) is not valid if \( t < 4 \).

Points of Discontinuity

Discontinuity in a function occurs at points where the function is not continuous. For a function to be continuous at a specific point, the function must be defined there, and its behavior should not "jump". Essentially, you could draw the graph of the function at that point without lifting your pencil off the paper.

In our case with \( g(t) = \sqrt{t - 4} \), t=3 falls outside the domain \([4, \infty)\), rendering \( g(3) \) undefined. Consequently, there is a break or discontinuity at t=3, since the function simply doesn't exist at this point. This showcases the importance of domain in analyzing continuity: if a point is outside a function’s domain, it is automatically a point of discontinuity.