Question

a. Let \(g(x)=2 x+3\) and \(h(x)=x^{3} .\) Consider the composite function \(f(x)=g(h(x))\). Find \(f^{-1}\) directly and then express it in terms of \(g^{-1}\) and \(h^{-1}\). b. Let \(g(x)=x^{2}+1\) and \(h(x)=\sqrt{x} .\) Consider the composite function \(f(x)=g(h(x))\). Find \(f^{-1}\) directly and then express it in terms of \(g^{-1}\) and \(h^{-1}\). c. Explain why if \(g\) and \(h\) are one-to-one, the inverse of \(f(x)=g(h(x))\) exists.

Short answer

Short Answer: If g and h are one-to-one functions, the composite function f(x) = g(h(x)) is also one-to-one and therefore invertible, meaning a unique inverse function exists. We can find this inverse function directly by solving for x in terms of y, or by expressing it in terms of the inverses of g and h.

Step-by-step solution

Find the composite function f(x)

First, we will substitute h(x) into g(x) to find f(x): \(f(x)=g(h(x))=2h(x)+3=2(x^3)+3=2x^3+3\)

Find the inverse of f(x) directly

Now, we need to find the inverse function \(f^{-1}(x)\). To do this, we replace f(x) with y and then swap x and y: \(y=2x^3+3 \Rightarrow x=2y^3+3\). Now, we need to solve for y: \(x=2y^3+3 \Rightarrow y^3=\frac{x-3}{2} \Rightarrow y=f^{-1}(x)=(\frac{x-3}{2})^{\frac{1}{3}}\)

Find the inverses of g(x) and h(x)

Now, let's find the inverses of the functions g(x) and h(x). For \(g(x)=2x+3\), we have \(y=2x+3 \Rightarrow x=2y+3 \Rightarrow y=g^{-1}(x)=\frac{x-3}{2}\). For \(h(x)=x^3\), we have \(y=x^3 \Rightarrow x=y^3 \Rightarrow y=h^{-1}(x)=x^{\frac{1}{3}}\).

Express f^{-1}(x) in terms of \(g^{-1}\) and \(h^{-1}\)

To express \(f^{-1}(x)\) in terms of \(g^{-1}\) and \(h^{-1}\), we need to find the inverse of \(f(x)=g(h(x))\) in the form \(f^{-1}(x)=h^{-1}(g^{-1}(x))\). Using the inverses we found above, we can express \(f^{-1}(x)\) as follows: \(f^{-1}(x)=h^{-1}(g^{-1}(x))=h^{-1}(\frac{x-3}{2})=(\frac{x-3}{2})^{\frac{1}{3}}\) Thus, we found the same result as in step 2, and proved that the inverse of the composite function can be expressed in terms of the inverses of the individual functions g and h. #b. Let \(g(x)=x^{2}+1\) and \(h(x)=\sqrt{x}\). Consider the composite function \(f(x)=g(h(x))\). Find \(f^{-1}\) directly and then express it in terms of \(g^{-1}\) and \(h^{-1}\).

Find the composite function f(x)

First, we will substitute h(x) into g(x) to find f(x): \(f(x)=g(h(x))=g(\sqrt{x})=x+1\)

Find the inverse of f(x) directly

Now, we need to find the inverse function \(f^{-1}(x)\). To do this, we replace f(x) with y and then swap x and y: \(y=x+1 \Rightarrow x=y+1\). Now, we need to solve for y: \(x=y+1 \Rightarrow y=f^{-1}(x)=x-1\)

Find the inverses of g(x) and h(x)

Now, let's find the inverses of the functions g(x) and h(x). For \(g(x)=x^{2}+1\), we have \(y=x^2+1 \Rightarrow x=y^2+1 \Rightarrow y=g^{-1}(x)=\sqrt{x-1}\). For \(h(x)=\sqrt{x}\), we have \(y=\sqrt{x} \Rightarrow x=y^2 \Rightarrow y=h^{-1}(x)=x^2\).

Express f^{-1}(x) in terms of \(g^{-1}\) and \(h^{-1}\)

To express \(f^{-1}(x)\) in terms of \(g^{-1}\) and \(h^{-1}\), we need to find the inverse of \(f(x)=g(h(x))\) in the form \(f^{-1}(x)=h^{-1}(g^{-1}(x))\). Using the inverses we found above, we can express \(f^{-1}(x)\) as follows: \(f^{-1}(x)=h^{-1}(g^{-1}(x))=h^{-1}(\sqrt{x-1})=(\sqrt{x-1})^2=x-1\) Thus, we found the same result as in step 2, and proved that the inverse of the composite function can be expressed in terms of the inverses of the individual functions g and h. #c. Explain why if \(g\) and \(h\) are one-to-one, the inverse of \(f(x)=g(h(x))\) exists. If both functions g and h are one-to-one, that means for each value in the domain of g and h, there is a unique value in their respective ranges. When we form the composite function f(x) = g(h(x)), the resulting function is also one-to-one, as it maps the unique values of h(x) to unique values in the range of g(x). Since the composite function is one-to-one, it is invertible. That implies that a unique inverse function exists for the composite function f(x) = g(h(x)), and we can find it by solving for x in terms of y or expressing it in terms of the inverses of g and h.

Key concepts

Composite Functions

When we talk about composite functions, we are referring to the idea of combining two functions into one. If you have two functions, say \( g(x) \) and \( h(x) \), then their composite is written as \( f(x) = g(h(x)) \). This means you first apply the function \( h \) to \( x \), and then you take that result and apply \( g \) to it.

Here are some key points to keep in mind about composite functions:

• The order in which you compose the functions matters. \( g(h(x)) \) is not the same as \( h(g(x)) \).
• Checking the domain of the composite function is important, as you need to ensure the output of \( h(x) \) falls within the domain of \( g(x) \).

Understanding composite functions lets you tackle problems in a step-by-step manner, first solving the inside function and then applying the result to the outer function.

Function Composition

Function composition involves creating a new function by feeding the output of one function to another. Think of it as linking processes where the output of one process becomes the input of another.

For instance, with functions \( g \) and \( h \), the composition \( f(x) = g(h(x)) \) effectively passes the result of \( h(x) \) into \( g \). This is crucial in finding inverses of composite functions.

• To reverse a composite function, \( f(x) \), you can use the inverses of the individual functions, \( g^{-1} \) and \( h^{-1} \), accordingly.
• The inverse of \( f(x) \) can be represented as \( f^{-1}(x) = h^{-1}(g^{-1}(x)) \), effectively undoing each function in reverse order of the composition.

Function composition is a powerful tool that allows for complex functions to be broken down and analyzed in manageable steps, facilitating a deeper understanding of their behavior and inverses.

One-to-One Functions

A one-to-one function is a function where each output is paired with exactly one unique input. This is a critical idea when dealing with function inverses because only one-to-one functions have inverses.

Here are some properties of one-to-one functions:

• They pass the horizontal line test. If any horizontal line intersects a function more than once, the function is not one-to-one.
• This property ensures that no two different elements in the domain map to the same element in the codomain.

When both \( g \) and \( h \) are known to be one-to-one, their composite \( f(x) = g(h(x)) \) will also be one-to-one.

This ensures that an inverse function for \( f(x) \) exists. You can confidently say that for each output of \( f \), there is one unique input, allowing you to reverse the function accurately.