/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Show that the function \(f\) def... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 73

Show that the function \(f\) defined by \(f(x)=m x+b\) is an odd function if and only if \(b=0\).

Short Answer

Expert verified
To show that \(f(x) = mx + b\) is an odd function if and only if \(b = 0\), we prove two statements: (1) If the function is odd, then \(b = 0\), and (2) If \(b = 0\), then the function is odd. For the first part, given \(f(-x) = -f(x)\), after simplification, we find \(b = 0\). For the second part, assuming \(b = 0\), we have \(f(x) = mx\) and \(f(-x) = -mx\). We can see that \(f(-x) = -f(x)\), and thus, if \(b = 0\), the function is odd. Therefore, \(f(x) = mx + b\) is an odd function if and only if \(b = 0\).

Step by step solution

01

Proving that if the function is odd, then b=0

We are given that f(x) is an odd function. Therefore, it satisfies the condition \(f(-x) = -f(x)\). Let's use the condition for f(x): \(f(-x) = m(-x) + b\) \(-f(x) = -(mx + b)\) Now set these two expressions equal to each other: \(m(-x) + b = -(mx + b)\) Let's simplify and solve for b.
02

Simplifying and solving for b

Expanding and simplifying the equation: \(-mx - b = -mx + b\) Now, isolate b by adding mx to both sides: \(0 = 2b\) And divide both sides by 2 to find b: \(b = 0\) Since we have proven that if the function is odd, then b=0, we can move on to the second part of the proof.
03

Proving that if b=0, then the function is an odd function

Suppose that \(b = 0\). We want to show that under this condition, the function satisfies the definition of an odd function, that is, \(f(-x) = -f(x)\). Let's plug in the function definition and b=0: \(f(x) = mx + 0 = mx\) \(f(-x) = m(-x) + 0 = -mx\) Now let's compare \(f(-x)\) and \(-f(x)\): \(f(-x) = -mx\) \(-f(x) = -(mx)\) We can see that under the condition \(b = 0\), both expressions are equal: \(-mx = -mx\) Therefore, if \(b = 0\), then the function is an odd function. In conclusion, the function \(f(x) = mx + b\) is an odd function if and only if \(b = 0\).

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Most popular questions from this chapter

For each of the functions \(f\) given. (a) Find the domain of \(f\). (b) Find the range of \(f\). (c) Find a formula for \(f^{-1}\). (d) Find the domain of \(f^{-1}\). (e) Find the range of \(f^{-1}\). You can check your solutions to part (c) by verifying that \(f^{-1} \circ f=I\) and \(f \circ f^{-1}=I\) (recall that \(I\) is the function defined by \(I(x)=x\) ). \(f(x)=x^{2}+8\), where the domain of \(f\) equals \((0, \infty)\).

Find functions \(f, g,\) and \(h,\) each simpler than the given function \(T,\) such that \(T=f \circ g \circ h .\) \(T(x)=\frac{4}{5+x^{2}}\)

Give an example to show that the product of two one-to-one functions is not necessarily a one-to-one function.

Suppose \(f\) is a function and a function \(g\) is defined by the given expression. (a) Write \(g\) as the composition of \(f\) and one or two linear functions. (b) Describe how the graph of \(g\) is obtained from the graph of \(f\). \( g(x)=3 f(x)-2\)

Suppose \(f\) and \(g\) are functions, each with domain of four numbers, with \(f\) and \(g\) defined by the tables below: $$\begin{array}{c|c}x & f(x) \\\\\hline 1 & 4 \\\2 & 5 \\\3 & 2 \\\4 & 3\end{array}$$ $$\begin{array}{c|c}x & g(x) \\\\\hline 2 & 3 \\\3 & 2 \\\4 & 4 \\\5 & 1\end{array}$$ Give the table of values for \(g^{-1} \circ f^{-1}\).
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