Question

Continuity of a Composite Function In Exercises \(67-72\) discuss the continuity of the composite function \(h(x)=f(g(x))\) $$ \begin{array}{l}{f(x)=\frac{1}{\sqrt{x}}} \\ {g(x)=x-1}\end{array} $$

Short answer

The composite function \(h(x) = \frac{1}{\sqrt{x-1}}\) is continuous for \(x > 1\).

Step-by-step solution

Define g(x)

First, let's define the function \(g(x)\). In this case, \(g(x) = x - 1\). It can be observed that this function is a linear equation and hence, is continuous for all real numbers.

Define the Composite Function

Next, define the composite function \(h(x)=f(g(x))\). The function \(h(x)\) simply means wherever we see an \(x\) in \(f(x)\), replace it with \(g(x)\). By doing this, we get, \(h(x) = f(g(x)) = \frac{1}{\sqrt{g(x)}} = \frac{1}{\sqrt{x-1}}\)

Analyze the Composite Function

Now, we need to discuss the continuity of \(h(x) = \frac{1}{\sqrt{x-1}}\). This function is continuous wherever \(x-1\) is positive, i.e., \(x > 1\). For \(x \leq 1\), the denominator will become zero or negative, and the function is undefined. Thus, \(h(x) = \frac{1}{\sqrt{x-1}}\) is continuous for \(x > 1\).

Key concepts

Composite Functions

Composite functions are formed when one function is applied after another. In mathematical terms, if you have two functions, say, f(x) and g(x), the composite function h(x) = f(g(x)) means you first apply g(x) to your input, and then apply f(x) to the result of g(x). In the given exercise, we have g(x) = x - 1 and the function we are 'plugging it into' is f(x) = 1/√x.

Understanding composite functions is crucial as they help us to perform more complex operations by combining simpler functions. This concept has wide applications across various fields such as computer science, engineering, and economics, often allowing us to model complicated situations with simpler building blocks.

Continuity in Calculus

Continuity in calculus is a fundamental concept that determines whether a function can be drawn without lifting the pencil off the paper. In essence, a function is continuous at a point if you can find its limit, and that limit equals the function's value at that point. For a composite function like h(x) = f(g(x)), its continuity depends on the continuity of both f(x) and g(x) at the relevant points.

In the case at hand, since g(x) = x - 1 is a linear function, it is continuous everywhere. However, f(x) = 1/√x is only continuous where its input, x, is positive. Therefore, for the composite function h(x), we must ensure that g(x) is not just continuous but also positive, leading us to the conclusion that h(x) is continuous for x > 1.

Linear Functions

Linear functions are arguably the simplest functions in calculus. They can be represented by the equation f(x) = mx + b, where m is the slope, indicating the steepness of the line, and b is the y-intercept, the point where the line crosses the y-axis.

They're known for their straightforward properties: they have a constant rate of change, they're continuous and differentiable everywhere, and they graph to straight lines. In our example, g(x) = x - 1 is a linear function, thus it's continuous across all real numbers, making it a predictable and stable component in composite functions.

Square Root Functions

Square root functions, represented as f(x) = √x, are important non-linear functions that only take non-negative numbers as inputs. This is because square roots of negative numbers are not defined in the set of real numbers.

The domain of square root functions is therefore [0, ∞), meaning that it starts at 0 and includes all positive numbers. Their continuity is limited by this domain as they are undefined for negative inputs. When dealing with composite functions involving square roots, careful attention must be paid to ensure the inner function never produces a negative result, as we've seen with the continuity analysis of h(x) = 1/√(x - 1).