Problem 119 - Precalculus Essentials

Question

114\. If \(f(x)=x^{2}-4\) and \(g(x)=\sqrt{x^{2}-4},\) then \((f \circ g)(x)=-x^{2}\) and \(\left(f^{\circ} g\right)(5)=-25\) 115\. There can never be two functions \(f\) and \(g\), where \(f \neq g\), for which \((f \circ g)(x)=(g \circ f)(x)\) 116\. If \(f(7)=5\) and \(g(4)=7,\) then \((f \circ g)(4)=35\) 117\. If \(f(x)=\sqrt{x}\) and \(g(x)=2 x-1,\) then \((f \circ g)(5)=g(2)\) 118\. Prove that if \(f\) and \(g\) are even functions, then \(f g\) is also an even function. 119\. Define two functions \(f\) and \(g\) so that \(f^{\circ} g=g \circ f\)

Step-by-Step Solution

Answer

1) Yes, \((f \circ g)(x) = -x^{2}\) and \((f \circ g)(5) = -25\). 2) Generally, it's not true that two different functions \(f\) and \(g\) satisfy \((f \circ g)(x)=(g \circ f)(x)\). 3) No, \((f \circ g)(4) = 5\), not \(35\). 4) Yes, \((f \circ g)(5) = g(2)\) is correct. 5) Yes, if \(f\) and \(g\) are even functions, the product function \(fg\) is also even. 6) One example is when both functions are the identity, \(f(x) = g(x) = x\), so \((f \circ g)(x) = (g \circ f)(x) = x\).

Step 1: Problem 114

This problem already provides the result, i.e., \((f \circ g)(x)=-x^{2}\) and \((f \circ g)(5)=-25\). These results indicate that for any input \(x\), the composed function \((f \circ g)(x)\) gives an output of \(-x^{2}\). As such, there is no calculation needed. The result can be confirmed by substitution: \(f(g(x))=f(\sqrt{x^{2}-4})=(\sqrt{x^{2}-4})^{2}-4=-x^{2}\). When \(x=5\), this becomes \(-25\).

Step 2: Problem 115

To prove this, consider arbitrary functions \(f(x) = a_1*x + b_1\) and \(g(x) = a_2*x + b_2\). Computing both \((f \circ g)(x)\) and \((g \circ f)(x)\), we have \(f(g(x)) = a_1*(a_2*x+b_2) + b_1\) and \(g(f(x)) = a_2*(a_1*x+b_1) + b_2\). Except in trivial cases (i.e., when \(a_1 = a_2\) and \(b_1 = b_2\), which implies \(f = g\)), these quantities are not equal.

Step 3: Problem 116

Given \(f(7)=5\) and \(g(4)=7\), it is asked for \((f \circ g)(4)\), which becomes \(f(g(4)) = f(7) = 5\). Thus, \((f \circ g)(4)=5\), not \(35\).

Step 4: Problem 117

For \((f \circ g)(5)\), substitute \(5\) into the function \(g\) to get \(g(5)=2*5-1=9\). Then, substitute this result into function \(f\), we get \(f(9)=\sqrt{9}=3\). For \(g(2)\), substitute \(2\) into function \(g\) to get \(g(2)=2*2-1=3\). Therefore, \((f \circ g)(5) = g(2)\).

Step 5: Problem 118

To prove this, we first recall that a function \(h(x)\) is even if \(h(x) = h(-x)\). Therefore, if \(f\) and \(g\) are even, then for any \(x\), we can see that \((fg)(x) = f(x)g(x) = f(-x)g(-x) = (fg)(-x)\). Therefore, \(fg\) is also an even function.

Step 6: Problem 119

One example of functions \(f\) and \(g\) where \(f^{\circ} g=g \circ f\) would be the identity function. Given \(f(x) = x\) and \(g(x) = x\), it's easy to see that \((f \circ g)(x) = f(g(x)) = f(x) = x\) and \((g \circ f)(x) = g(f(x)) = g(x) = x\). Thus for these functions, \(f^{\circ} g=g \circ f\).

Key Concepts

Even Functions

Let's explore the fascinating concept of even functions. These are special types of functions that have symmetry about the y-axis. In mathematical terms, a function is considered even if for every number in the function's domain, the condition \(f(x) = f(-x)\) holds true. Imagine reflecting a shape across the y-axis; if the shape looks unchanged, the function of that shape is what we call an 'even' function.

An excellent example of an even function is \(f(x) = x^2\), which keeps its form whether you input positive or negative values of 'x'. This property is crucial in understanding problem 118 from our exercise, where proving that the product of two even functions remains an even function involves utilizing the definition of evenness. To validate this, we simply show that \(f(x)g(x) = f(-x)g(-x)\), hence maintaining the characteristic symmetry of even functions.

Function Identities

Function identities refer to a set of functions that possess certain invariance under specific operations, much like the identity property in algebra where \(a + 0 = a\) and \(a * 1 = a\). The most common function identity in mathematics is the identity function, denoted as \(f(x) = x\).

In our exercise, problem 119 showcases an example where two functions, when composed, result in an identity function. By setting both \(f(x)\) and \(g(x)\) as the identity function, \(f \circ g\) and \(g \circ f\) both yield the identity function, demonstrating the interchangeable nature and the true essence of function identities. This shows that certain function operations can preserve the identity relationship, an important concept when considering function composition and transformations.

Composite Functions

Composite functions are the products of a unique function operation known as function composition. This procedure involves applying one function to the results of another function. The notation \(f \circ g\) represents the composition of functions 'f' and 'g', where \(f \circ g)(x) = f(g(x))\).

Exercise problems 114 through 117 dive into the compositions of various functions, where the output of function 'g' becomes the input for function 'f'. For example, \( (f \circ g)(5) \) gives us a particular value after substituting '5' into 'g' first, then taking the result into 'f'. Understanding how to compose functions accurately is vital, as it is a foundational concept in advanced mathematics, particularly in calculus where it is used to find derivatives of composite functions.

Function Operations

When we talk about function operations, we're referring to various ways in which functions can be combined. The most common operations include addition, subtraction, multiplication, and division of functions, alongside composition. Each operation has its own rules and properties that dictate the outcome of combining functions.

For instance, when adding two functions, \(f(x)\) and \(g(x)\), their sum \(h(x) = f(x) + g(x)\) consists of adding the outputs of each function at any given 'x'. A more complex operation is function composition, displayed in several exercise problems, which involves applying one function to the result of another, as seen in \(f \circ g\). The key in mastering function operations lies in understanding both the individual function behaviors and how they interact when combined.