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Chapter 1: Problem 30

Suppose \(g\) is the function whose domain is the interval \([-2,2],\) with \(g\) defined on this domain by the formula $$ g(x)=\left(5 x^{2}+3\right)^{7777} $$ Explain why \(g\) is not a one-to-one function.

Short Answer

Expert verified
The function \( g(x) = (5x^2 + 3)^{7777} \) is not one-to-one because it has an even function (h(x) = 5x^2 + 3) inside an odd power (7777). As a result, g(x) = g(-x) for all x within the domain. For example, g(1) = g(-1), meaning g(x) has at least one pair of distinct inputs (x1, x2) with the same output, which goes against the definition of a one-to-one function.

Step by step solution

01

Understand the definition of a one-to-one function

A function is considered one-to-one (or injective) if for every pair of distinct inputs within its domain, the outputs are distinct too. Mathematically, a function g is one-to-one if and only if: $$ \forall x_1, x_2 \in [-2, 2],\ \ x_1 \ne x_2 \Rightarrow g(x_1) \ne g(x_2) $$ To prove that the function g(x) is not one-to-one, we need to find at least one pair (x1, x2) such that x1 is not equal to x2, but g(x1) is equal to g(x2).
02

Analyse the formula of g(x)

g(x) is defined as: $$ g(x) = (5x^2 + 3)^{7777} $$ First, notice that the function g(x) is always non-negative, because both the square function \( x^2 \) and the power function \( ( \cdot )^{7777} \) are non-negative for all x. Moreover, since the exponent is an odd number (7777), the function g(x) is strictly positive for all x 鈭� [-2, 2]. Secondly, observe that the function inside the power, h(x) = 5x^2 + 3, is an even function. This means that h(x) satisfies the property: $$ h(x) = h(-x) $$ For all x within the domain. Notice that g(x) has an even function (h(x)) inside an odd power (7777). We can use this fact to find a pair of x1 and x2 with the same output.
03

Find a pair of x1 and x2 with the same output

Since h(x) = h(-x), this means that g(x) = g(-x) for all x within the domain. We can find a pair of x1 and x2 with the same output by using this fact. For instance, let x1 = 1 and x2 = -1; then $$ g(x_1) = g(1) $$ and $$ g(x_2) = g(-1) $$ are equal because of the properties we have observed in Step 2. Therefore, we have $$ g(1) = g(-1) $$ which means that the function g(x) is not one-to-one.

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Most popular questions from this chapter

Assume that \(f\) is the function defined by $$ f(x)=\left\\{\begin{array}{ll} 2 x+9 & \text { if } x<0 \\ 3 x-10 & \text { if } x \geq 0. \end{array}\right. $$ Evaluate \(f(|x-5|+2)\).

Show that the composition of two one-to-one functions is a one-to-one function. [Here you need to assume that the two functions have range and domain such that their composition makes sense.]

Consider the function \(h\) whose domain is the interval [-3,3] , with \(h\) defined on this domain by the formula $$ h(x)=(3+x)^{2} $$ Does \(h\) have an inverse? If so, find it, along with its domain and range. If not, explain why not.

True or false: If \(f\) is an even function whose domain is the set of real numbers and a function \(g\) is defined by $$ g(x)=\left\\{\begin{array}{ll} f(x) & \text { if } x \geq 0 \\ -f(x) & \text { if } x<0 \end{array}\right. $$ then \(g\) is an odd function. Explain your answer.

Give an example of a function \(f\) whose domain is the set of real numbers and such that the values of \(f(-1), f(0)\), and \(f(2)\) are given by the following table: $$ \begin{array}{r|r} {r|} {x} & {c} {f(x)} \\ \hline-1 & \sqrt{2} \\ 0 & \frac{17}{3} \\ 2 & -5 \end{array} $$
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