91Ó°ÊÓ

When you decompose a function into two smaller functions, the inner function, denoted as \( h(x) \), is the first to process an input value. It operates inside, transforming the initial input before the outer function gets involved. To understand this, let's look at part (a) of our exercise, where we have the function \( f(x) = x^2 + 1 \).

Here, the inner function can be chosen as \( h(x) = x^2 \). This choice is strategic because it conveniently isolates the most basic operable part of the function \( f(x) \), which is the \( x^2 \) expression.

For part (b), the function \( f(x) = \frac{1}{x-3} \) is handled by selecting the inner function \( h(x) = x - 3 \). This effectively separates the \( x - 3 \) operation required inside the composite function. The inner function is crucial as it simplifies the job for the outer function, preparing the input in a specific way.

Once the input is transformed by the inner function, the outer function \( g(x) \) completes the composition by processing the result from the inner function. The outer function takes this output and transforms it further to give the final function.

In our exercise, for part (a), we identified the outer function as \( g(x) = x + 1 \). This means after the input \( x \) is squared by the inner function \( h(x) = x^2 \), the outer function adds 1, resulting in the complete expression \( x^2 + 1 \).

Similarly, in part (b), the outer function is set as \( g(x) = \frac{1}{x} \). It receives the output from \( h(x) = x - 3 \) and processes it so that we have \( \frac{1}{x - 3} \). The outer function finalizes the transformation, completing the composition to match our original function \( f(x) \).

Function verification is a vital step to ensure that the decomposition is correctly performed. It involves substituting the inner function into the outer function to verify if the original function \( f(x) \) is recovered.

Let's take part (a) of our exercise as an example. With the composition \( f(x) = g(h(x)) \), our inner function \( h(x) \) is \( x^2 \) and our outer function \( g(x) \) is \( x + 1 \). When we substitute \( h(x) \) into \( g(x) \), it becomes \( g(h(x)) = (x^2) + 1 = x^2 + 1 \), thus confirming the original function.

Similarly, for part (b), substituting \( h(x) = x - 3 \) into \( g(x) = \frac{1}{x} \) verifies our composition, as \( g(h(x)) = \frac{1}{x - 3} \).

Verification assures the accuracy of the decomposition and is an essential check before concluding that the functions have been correctly composed.