Question

Find functions \(f, g\), and h such that \(F=f \circ g \circ h .\) (Note: The answer is not unique.) a. \(F(x)=\sqrt{1-\sqrt{x}}\) b. \(F(x)=\sin ^{3}(2 x+3)\)

Short answer

For function \(F(x)=\sqrt{1-\sqrt{x}}\), we have \(f(x)=\sqrt{x}\), \(g(x)=1-x\), and \(h(x)=\sqrt{x}\) such that \(F(x)=f \circ g \circ h (x)\). For function \(F(x)=\sin^{3}(2x+3)\), we have \(f(x)=x^3\), \(g(x)=\sin(x)\), and \(h(x)=2x+3\) such that \(F(x)=f \circ g \circ h (x)\).

Step-by-step solution

Identify inner function h

Observe that the innermost operation is taking a square root. Let's consider \(h(x)=\sqrt{x}\).

Identify the middle function g

After applying h, we have \(\sqrt{1-\sqrt{x}} = \sqrt{1-h(x)}\). The next operation is subtraction, let's consider \(g(x)=1-x\).

Identify the outer function f

Finally, we have \(\sqrt{1-\sqrt{x}} = \sqrt{g(h(x))}\), which implies that the outer function is taking the square root of its input, so let \(f(x)=\sqrt{x}\). Thus, we have functions \(f(x)=\sqrt{x}\), \(g(x)=1-x\), and \(h(x)=\sqrt{x}\) such that \(F(x)=f \circ g \circ h (x)\). b. \(F(x)=\sin^{3}(2x+3)\)

Identify inner function h

Observe that the innermost operation is adding 3 to 2x. Let's consider \(h(x)=2x+3\).

Identify the middle function g

After applying h, we have \(\sin^{3}(2x+3)=\sin^{3}(h(x))\). The next operation is taking the sine of its input, so let \(g(x)=\sin(x)\).

Identify the outer function f

Finally, we have \(\sin^{3}(2x+3)=(\sin(h(x)))^3\), which implies that the outer function is taking the cube of its input, so let \(f(x)=x^3\). Thus, we have functions \(f(x)=x^3\), \(g(x)=\sin(x)\), and \(h(x)=2x+3\) such that \(F(x)=f \circ g \circ h (x)\).

Key concepts

Function Composition

Function composition is akin to a culinary recipe where ingredients are combined in a specific sequence to create a final dish. In the realm of calculus, function composition involves combining two or more functions in a sequence to form a new function. This can be written as \( F = f \circ g \circ h \) where \( f, g, \text{and} h \) are individual functions and \( \circ \) represents composition. Imagine \( g \) as a chef preparing a base ingredient, then \( f \) takes that prepared ingredient to finish the recipe. The beauty lies in the fact that the result of one function becomes the input for the next. This concept is not only essential for understanding complex mathematical relationships but also for solving problems that involve layered processes or procedures. \(
\) When using function composition, remember that order matters. Reversing the sequence of functions can lead to completely different results, just as altering recipe steps can yield a different dish. In the provided examples, by carefully selecting functions \( h, g, \text{and} f \) you can construct \( F \) in such a way that the operations properly layer to produce the desired outcome.

Inner Function

The inner function is the foundation of our composed function—like the base layer of a cake that everything else is built upon. In the context of the exercise, the inner function, \( h(x) \), is identified as the first operation to apply to the input \( x \). It is 'inner' because its output is used as the input for the next function in the composition sequence. \(
\) For example, \( h(x)=\sqrt{x} \) is the foundation in the equation \( F(x)=\sqrt{1-\sqrt{x}} \) since we must first compute \( \sqrt{x} \) before we can subtract and take another square root. This step-by-step approach, focusing on the inner function first, simplifies complex equations, making them easier to understand and solve. Recognizing and correctly identifying the inner function is a pivotal step in the process of decomposing a composed function.

Outer Function

Opposite to the inner function, the outer function is like the frosting on our layered cake—it's applied last. After we've performed the operations defined by the inner and any middle functions, the outer function, \( f(x) \) in our example, is applied to the result we've obtained. In the equation \( F(x)=\sqrt{1-\sqrt{x}} \) the outer function is \( f(x)=\sqrt{x} \) which operates on the output of our previous workings. \(
\) The correct identification and application of the outer function determine the final form of the composed function. Just like a cake without frosting feels incomplete, overlooking the outer function would lead to an incomplete or incorrect solution to the problem.

Operations on Functions

Operations on functions are the mathematical equivalent of different cooking techniques utilized in a recipe. Just as you might chop, boil, or fry ingredients, mathematical operations—such as addition, subtraction, multiplication, division, exponentiation, and trigonometric functions—are applied to functions. In function composition, these operations are not performed separately but in a specific order as dictated by the composition. \(
\) The steps in our exercise illustrate how operations transform the input progressively through each function. From adding constants, taking roots, to performing trigonometric operations, each has a specific place in the sequence. Understanding how and why these operations affect outputs is fundamental to tackling calculus problems effectively. Remember to think of operations not just as isolated actions but as vital elements of a larger, ordered process that culminates in the final function.