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Chapter 1: Problem 29

Suppose \(f\) is the function whose domain is the set of real numbers, with \(f\) defined on this domain by the formula $$ f(x)=|x+6| $$ Explain why \(f\) is not a one-to-one function.

Short Answer

Expert verified
The function \(f(x) = |x+6|\) is not one-to-one because there are two different input values that produce the same output value. Specifically, we found that when x = 2 and x = -14, both cases yield f(x) = 8. Therefore, the function does not meet the requirement of a one-to-one function, where each output value must correspond to exactly one input value.

Step by step solution

01

Understanding of one-to-one functions

In a one-to-one function, each output value corresponds to exactly one input value. In other words, if f(x1) = f(x2), then x1 must equal x2.
02

Find values of x that produce the same output

For the function f(x) = |x+6|, let's try finding two different values of x that yield the same output for f(x). Consider the case when x = -6 and x = -8.
03

Calculate f(x) for both values

Now let's calculate the output values for the function f(x) for both x = -6 and x = -8: For x = -6: \(f(-6) = |(-6) + 6| = |-6 + 6| = |0| = 0\) For x = -8: \(f(-8) = |(-8) + 6| = |-8 + 6| = |-2| = 2\) However, in this case, we see that f(-6) 鈮� f(-8). Let's try another set of values for x.
04

Find another set of values for x that produce the same output

Let's try x = -6 and x = -10 this time.
05

Calculate f(x) for the new values

Calculate the output values for the function f(x) for both x = -6 and x = -10: For x = -6 (same as before): \(f(-6) = |(-6) + 6| = |-6 + 6| = |0| = 0\) For x = -10: \(f(-10) = |(-10) + 6| = |-10 + 6| = |-4| = 4\) Since f(-6) 鈮� f(-10), let's try another set of values for x.
06

Find another set of values for x that produce the same output

Let's try x = 2 and x = -14 this time.
07

Calculate f(x) for the third set of values

Calculate the output values for the function f(x) for both x = 2 and x = -14: For x = 2: \(f(2) = |(2) + 6| = |2 + 6| = |8| = 8\) For x = -14: \(f(-14) = |(-14) + 6| = |-14 + 6| = |-8| = 8\) Now, we see that f(2) = f(-14), which shows that there are two different input values (2 and -14) that produce the same output value (8).
08

Conclusion

Since we found two different values of x, specifically x = 2 and x = -14, for which f(2) = f(-14), we can conclude that the function f(x) = |x+6| is not a one-to-one function.

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Most popular questions from this chapter

Suppose $$ h(x)=2+\sqrt{\frac{1}{x^{2}+1}} $$ (a) If \(g(x)=\frac{1}{x^{2}+1},\) then find a function \(f\) such that \(\bar{h}=f \circ g\) (b) If \(g(x)=x^{2},\) then find a function \(f\) such that \(h=f \circ g\).

In Exercises \(37-40,\) find functions \(f\) and \(g,\) each simpler than the given function \(h\), such that \(\boldsymbol{h}=\boldsymbol{f} \circ \mathrm{g}\) $$ h(x)=\frac{2}{3+\sqrt{1+x}} $$

(a) True or false: The product of an even function and an odd function (with the same domain) is an odd function. (b) Explain your answer to part (a). This means that if the answer is "true", then explain why the product of every even function and every odd function (with the same domain) is an odd function; if the answer is "false", then give an example of an even function \(f\) and an odd function \(g\) (with the same domain) such that \(f g\) is not an odd function.

Suppose h is defined by \(h(t)=|t|+1\). What is the range of \(h\) if the domain of \(h\) is the interval [-8,2]\(?\)

A constant function is a function whose value is the same at every number in its domain. For example, the function \(f\) defined by \(f(x)=4\) for every number \(x\) is a constant function. Suppose \(f\) is an even function and \(g\) is an odd function such that the composition \(f \circ g\) is defined. Show that \(f \circ g\) is an even function.
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