Negative reciprocal – Explanation and Examples

Negative reciprocal may sound complicated, but once we’ve understood its concept, you’ll see how easy it is to apply and find a number’s negative reciprocal. Why don’t we dissect the two words?

Negative and reciprocal – this means that a number’s negative reciprocal is the result of multiplying the number’s reciprocal by $\mathbf{-1}$.

As straightforward as its definition, negative reciprocals have a wide range of applications that include finding perpendicular slopes and modeling real-world applications that make use of inverse relationships.

What is a negative reciprocal?

When dealing with negative reciprocals, we’ll first recall what these two words represent in math: negative and reciprocal.

$ \boldsymbol{\dfrac{a}{b} \rightarrow – \dfrac{b}{a} }$

We’ll slowly break down this form, and by the end of this article, you’ll definitely be able to understand what this represents.

Reciprocal

The reciprocal of a number or a function is the value or expression that results from reversing the numerator and denominator’s places.

$ \boldsymbol{\dfrac{a}{b}\rightarrow \dfrac{b}{a} }$

Reciprocals are considered multiplicative inverses since they will always be 1 when we multiply a number by its reciprocal.

$ \dfrac{a}{b} \cdot \dfrac{b}{a}=1$

Master your knowledge on reciprocals here.

Negative of a Number (or a Function)

The negative of a number or a function is the result of a number being multiplied by -1. Let’s say we have a fraction, $\dfrac{b}{a}$, its negative counterpart will be $-\dfrac{b}{a}$.

$ \boldsymbol{-1 \cdot \dfrac{b}{a} = -\dfrac{b}{a}}$

Learn more about negative numbers here.

When we combine these two concepts, we’ll have the negative reciprocal of a number. This means that negative reciprocals result from us taking the reciprocal of a number then find the negative value of the result.

Hence, we have $ \boldsymbol{\dfrac{a}{b} \rightarrow – \dfrac{b}{a} }$.

How to find negative reciprocal?

Now that we understand what negative reciprocals represent, how do we manipulate different forms of expressions to have their negative reciprocals?

  • Always begin by switching the places of the numerator and the denominator of the fraction.
  • Once we have the reciprocal, multiply the result by $\mathbf{-1}$.

We’ve created quick guides you can note when working with different types of numbers and expressions.

Let’s start by learning to find the negative reciprocal of a fraction, $\dfrac{a}{b}$, where $b \neq 0$.

What if we’re working with rational functions  such as $\dfrac{p(x)}{q(x)}$? We apply the same process as we did with fractions.

The rational function and its negative reciprocal will only be valid when $p(x) \neq 0$ and $q(x) \neq 0$.

What if we’re working with whole numbers then? We begin by expressing the whole number as a fraction with $1$ as its denominator. Let’s say we have $m$ as the whole number, we start by expressing it as $\dfrac{m}{1}$ and follow the same process.

A similar process applies for functions such as $f(x)$ that are not rational functions.

Take note that for the negative reciprocals to exist, both $m$ and $f(x)$ must not be equal to $0$.

Excited to try out problems involving negative reciprocals? First, let’s go ahead and summarize what we’ve learned so far about negative reciprocals.

Summary of reciprocal definition and properties

  • This expression represents what occurs in finding the negative reciprocals: $ \boldsymbol{\dfrac{a}{b} \rightarrow – \dfrac{b}{a} }$.
  • When given a whole number or a function that is not rational, begin by expressing the given as a fraction of 1.
  • It is only possible for a constant or a function to have a negative reciprocal when both its numerator and denominator are not equal to $0$.
  • A slope of a perpendicular line makes use of negative reciprocals.

That’s it. Make sure to keep these pointers in mind when solving the problems below.

Example 1

Complete the table below by finding the respective negative reciprocals of the following.

Original ValueNegative Reciprocal
$\dfrac{1}{2}$ 
$-\dfrac{2}{3}$ 
$9$ 
$- 4\dfrac{1}{7}$ 

Solution

When finding the negative reciprocal, we begin by switching the fraction’s numerator and denominator places. Let’s work on the first two items first: $\dfrac{1}{2}$ and $-\dfrac{2}{3}$.

Hence, their reciprocals are $\dfrac{2}{1}$ and $-\dfrac{3}{2}$.

For each value, multiply $-1$ to find the corresponding negative reciprocal.

  • $-1 \cdot \dfrac{2}{1} = -2$
  • $-1 \cdot -\dfrac{3}{2}=\dfrac{3}{2}$.

We’ll actually apply the same process for the last two rows, but let’s first make sure we rewrite them in fraction form. The whole number $9$ can be written as $\dfrac{9}{1}$ and the mixed number $- 4\dfrac{1}{7}$ can be written as $-\dfrac{29}{7}$.

Once we have them in fraction forms, we can now switch the places of their corresponding numerators and denominators then multiplying the respective result by $-1$.

  • $\begin{aligned}\dfrac{9}{1} \rightarrow \dfrac{1}{9} \rightarrow -\dfrac{1}{9} \end{aligned}$
  • $\begin{aligned} -\dfrac{29}{7} \rightarrow \dfrac{-7}{29} \rightarrow \dfrac{7}{29} \end{aligned}$

Hence, we have the completed table as shown below.

Original ValueNegative Reciprocal
$\dfrac{1}{2}$$-2$
$-\dfrac{2}{3}$$\dfrac{3}{2}$
$9$$-\dfrac{1}{9}$
$- 4\dfrac{1}{7}$$-\dfrac{7}{29}$

Example 2

Let $h(x)$ be the negative reciprocal of $f(x)$ for each of the following functions. Find the $h(x)$. What are the restrictions for $x$ in each case?

a. $f(x) = \dfrac{1}{x – 1}$

b. $f(x) = \dfrac{2}{3(x +2)}$

c. $f(x) = x^2 – 3x – 54$

Solution

We apply the same process when finding the negative reciprocals of functions.

a. This means that we begin by switching the places of $1$ and $x – 1$ to find the reciprocal of $f(x)$. We then multiply the result by $-1$.

$\begin{aligned}h(x)&=-1\cdot \dfrac{x-1}{1}\\&=-1\cdot x – 1\\&\mathbf{-x + 1} \end{aligned}$

Since $h(x)$ is a linear expression, it has not restrictions. The function $f(x)$, however, must not have $x – 1 = 0$, so $\mathbf{x \neq 0}$.

b. We apply the same process from a. Hence, we have $h(x)$ as shown below.

$\begin{aligned}h(x)&=-1\cdot \dfrac{3(x+2)}{2}\\&=-1\cdot \dfrac{3x+6}{2}\\&=\mathbf{-\dfrac{3x+6}{2}} \end{aligned}$

The function $h(x)$ has a constant as its denominator, so it has no restrictions for $x$. The function $f(x)$, however, can’t have $3(x + 2) = 0$, so $\mathbf{x \neq -2}$.

c. Express $f(x)$ as a fraction by having $1$ as its denominator, so $f(x) = \dfrac{ x^2 – 3x – 54}{1}$. Now, apply the same process to find the negative reciprocal, $h(x)$.

$\begin{aligned}h(x)&=-1\cdot \dfrac{1}{x^2-3x-54}\\&=\mathbf{-\dfrac{1}{x^2-3x-54}}\\ \end{aligned}$

Since $f(x)$ is a polynomial, it has no restrictions for $x$. Its negative reciprocal, however, can’t have zero in its denominator. We can find the restrictions for $h(x)$ by finding the values where $ x^2 – 3x – 54$ is zero.

$ \begin{aligned} x^2 -3x – 54&=0\\(x – 9)(x + 6)&=0\\x&=9\\x&-6\end{aligned}$

This means that for $h(x)$ to be valid, $\mathbf{x \neq \{-6,9\}}$.

Example 3

The graph of the linear function, $f(x)$, is perpendicular to the graph of $h(x)$, which is also linear function. If $f(x)$ has a slope of $-\dfrac{2}{3}$, what is the slope of $h(x)$?

Solution

As we have mentioned in the discussion, finding negative reciprocals is crucial when finding the slopes of perpendicular lines.

Since we have the slope of $f(x)$, we can find the slope of $h(x)$ by finding the negative reciprocal of $-\dfrac{2}{3}$.

$\begin{aligned}m_\perp &= -1 \cdot-\dfrac{3}{2}\\&=\dfrac{3}{2} \end{aligned}$

This means that the slope of $h(x)$ is $\dfrac{3}{2}$ for it to be perpendicular to $f(x)$.

Example 4

The negative reciprocal of $f(x)$ is $\dfrac{x^2 – 2}{x – 5}$. What is the expression for $f(x)$?

Solution

This time, we’re given the negative reciprocal. We need to find the expression for $f(x)$ by reversing the steps:

  • We start by multiplying $-1$ back on the negative reciprocal to reverse the changes in the sign.
  • Switch the places of the negative reciprocal’s numerator and denominator.

$\begin{aligned}f(x)&=-1\cdot \dfrac{x-5}{x^2-2}\\&=\dfrac{-x+5}{x^2-2} \end{aligned}$

This means that $\mathbf{f(x) =\dfrac{-x+5}{x^2-2}}$.

Notice something about the steps? They are actually the same process because the negative reciprocal of a function’s negative reciprocal will be $\mathbf{f(x)}$.

Example 6

If a given number is twenty-seven times larger than its negative reciprocal’s square, find the number.

Solution

Let $n$ be the number we’re looking for, so its negative can be expressed as $-\dfrac{1}{n}$. Set up the equation representing the situation.

$n=27\cdot\left(-\dfrac{1}{n} \right )^2 $

Simplify this equation by multiplying both sides of the equation by $n^2$ and taking the cube root of both sides of the equation.

$\begin{aligned}n&=\dfrac{27}{n^2}\\n^3&=27\\\sqrt[3]{n^3} &=\sqrt[3]{27}\\n&=3  \end{aligned}$

This means that for the number to satisfy the condition, it must be equal to 3.

Practice Questions

1. Complete the table below by finding the respective negative reciprocals of the following.

Original ValueNegative Reciprocal
$\dfrac{1}{5}$ 
$-\dfrac{6}{11}$ 
$-12$ 
$2\dfrac{3}{8}$ 

2. Let $h(x)$ be the negative reciprocal of $f(x)$ for each of the following functions. Find the $h(x)$. What are the restrictions for $x$ in each case?

a. $f(x) = \dfrac{2}{3x – 5}$
b. $f(x) = \dfrac{x}{2(x – 3)}$
c. $f(x) = x^2 – 7x – 30$
d. $f(x) = 1 + \dfrac{1}{x – 2}$

3. True or False? The reciprocal of the negative reciprocal of a function is equal to the function itself.
4. The graph of the linear function, $f(x)$, is perpendicular to the graph of $h(x)$, which is also linear function. If $f(x)$ has a slope of $-2\dfrac{1}{5}$, what is the slope of $h(x)$?
5. The negative reciprocal of $f(x)$ is $\dfrac{x^2 – 2}{x – 5}$. What is the expression for $f(x)$?
6. Let $h(x)$ be the negative reciprocal of $f(x)$.
a. What is the expression of $h(x)$ given that $f(x) = \dfrac{4x – 3}{2}$?
b. What are the restrictions for $x$ so that both $f(x)$ and $h(x)$ exist?
c. Use your knowledge in graphing reciprocal functions to graph $h(x)$. Include the vertical and horizontal asymptotes.
7. If a given number is sixty-four times larger than its negative reciprocal’s square, find the number.

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