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Chapter 5: Problem 94

Let \(f\) and \(g\) be one-to-one functions. Prove that (a) \(f \circ g\) is one-to-one. (b) \((f \circ g)^{-1}(x)=\left(g^{-1} \circ f^{-1}\right)(x)\)

Short Answer

Expert verified
The composition of two one-to-one functions is also one-to-one, and the inverse of the composition of two functions is equal to the composition of the inverses of the functions.

Step by step solution

01

Understand one-to-one functions and their composition

A function is said to be one-to-one (or injective) if every element of the function's domain corresponds to exactly one unique element of the function's range. In other words, if \(f(x1) = f(x2)\), then \(x1 = x2\). The composition of functions, denoted by '鈭�', combines two functions into a new function, where the output of one function becomes the input of the other.
02

Proving that the composition of one-to-one functions is also one-to-one

To prove this, we must show that if \(f(g(x1)) = f(g(x2))\), then \(x1 = x2\). If \(f(g(x1)) = f(g(x2))\), since \(f\) is one-to-one, we must have \(g(x1) = g(x2)\). And again, since \(g\) is also one-to-one, it follows that \(x1 = x2\). This proves that the composition of two one-to-one functions is also one-to-one.
03

Proving that the inverse of the composition of two functions is equal to the composition of the inverses of the functions

To prove this, we must show that \((f鈭榞)^{-1}(x) = (g^{-1}鈭榝^{-1})(x)\) for all \(x\). Let \(y = f(g(x))\), then \(f^{-1}(y) = g(x)\). Hence, \(g^{-1}(f^{-1}(y)) = x\). This means that \((g^{-1}鈭榝^{-1})(y)=x\), but \(y = f(g(x)) = (f鈭榞)(x)\). Therefore, \((g^{-1}鈭榝^{-1})(f(g(x))) = x\), proving that \((f鈭榞)^{-1}(x) = (g^{-1}鈭榝^{-1})(x)\) for all \(x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

One-to-One Functions
A one-to-one function, also referred to as an injective function, is a type of function where each element in the input set (domain) maps to a unique element in the output set (range). This means that no two different inputs produce the same output. Mathematically, a function \( f \) is one-to-one if whenever \( f(x_1) = f(x_2) \), it implies \( x_1 = x_2 \).

Here's why one-to-one functions are special:
  • They ensure that different inputs do not collapse into the same output, maintaining distinct results for each unique input.
  • This distinctness is crucial when considering the function's inverse, aiding in defining a proper functional relationship in the opposite direction.
Understanding one-to-one functions is foundational for tackling other advanced concepts like inverse functions and functional composition.
Inverse Functions
Inverse functions essentially "undo" the actions of the original function. If you have a function \( f \) and its inverse, denoted as \( f^{-1} \), applying \( f \) and then \( f^{-1} \) (or in reverse order) will bring you back to the starting point. Formally, if \( f(a) = b \), then \( f^{-1}(b) = a \).

For a function \( f \) to have an inverse, it must be one-to-one:
  • The one-to-one property ensures that each output can be traced back to exactly one input, making reverse mapping possible.
  • Being able to find an inverse makes it easier to solve equations and determine original inputs given outputs.
Exploring inverse functions leads to deeper insights into their behavior and the significance of the one-to-one requirement.
Function Composition
Function composition involves creating a "new" function by combining two existing functions. Denoted by \( f \circ g \), it represents applying function \( g \) first and then applying function \( f \) to the result of \( g \). This leads to a composite function where the output of \( g \) becomes the input for \( f \).

Key aspects of function composition include:
  • The process is sequential, meaning the order matters. \( f \circ g \) can yield a very different result compared to \( g \circ f \).
  • Function composition is a powerful tool in constructing complex transformations from simpler functions.
  • When both functions involved are one-to-one, the composition \( f \circ g \) will also be one-to-one, preserving the distinctiveness of input-output pairs.
This concept is widely used in various aspects of mathematics and computer science.
Injective Function Proof
Proving that a function is injective involves demonstrating that it respects the one-to-one property. For any function \( h \), if \( h(x_1) = h(x_2) \), then it must follow that \( x_1 = x_2 \). This ensures that \( h \) maps each element in the domain uniquely to an element in the range.

To prove injection through composition:
  • If \( f \) and \( g \) are injective, and we know \( f(g(x_1)) = f(g(x_2)) \), then injectivity requires:\( g(x_1) = g(x_2) \), since \( f \) being injective implies equal results for equal inputs.
  • This further implies \( x_1 = x_2 \), because \( g \) is injective by the same logic.
  • Thus, \( f \circ g \) must retain the one-to-one characteristic, proving the composite function is injective as well.
These proofs help solidify the distinct relationships between composite functions and their injective nature.

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Most popular questions from this chapter

Use a computer algebra system to find the linear approximation \(P_{1}(x)=f(a)+f^{\prime}(a)(x-a)\) and the quadratic approximation \(P_{2}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}\) of the function \(f\) at \(x=a\) . Sketch the graph of the function and its linear and quadratic approximations. \(f(x)=\arctan x, \quad a=1\)

Using a Right Triangle Show that \(\arctan (\sinh x)=\arcsin (\tanh x)\)

Use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point. \(\arcsin x+\arcsin y=\frac{\pi}{2}, \quad\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\)

Use a computer algebra system to find the linear approximation \(P_{1}(x)=f(a)+f^{\prime}(a)(x-a)\) and the quadratic approximation \(P_{2}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}\) of the function \(f\) at \(x=a\) . Sketch the graph of the function and its linear and quadratic approximations. \(f(x)=\arctan x, \quad a=0\)

Area In Exercises \(125-128\) , find the area of the region bounded by the graphs of the equations. Use a graphing utility to graph the region and verify your result. $$ y=x e^{-x^{2} / 4}, y=0, x=0, x=\sqrt{6} $$
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