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Chapter 2: Problem 94

Explain why \(g(x)=\frac{1}{-x+1}\) can be graphed by shifting the graph of\(f(x)=\frac{1}{x}\) one unit to the left and reflecting across the \(y\)-axis, orby shifting the graph of \(f\) one unit to the right and reflecting across the\(x\)-axis.

### Short Answer

Expert verified

Shift the graph of \( f(x) = \frac{1}{x} \) one unit to the left and reflect across the y-axis, or shift it one unit to the right and reflect across the x-axis.

## Step by step solution

01

## Understand the problem

We need to demonstrate why the function \( g(x) = \frac{1}{-x+1} \) can be graphed by either shifting the graph of \( f(x) = \frac{1}{x} \) one unit to the left and reflecting it across the \( y \)-axis, or by shifting \( f(x) \) one unit to the right and reflecting it across the \( x \)-axis.

02

## Rewrite the function

Start by rewriting \( g(x) \): \[ g(x) = \frac{1}{-x+1} = \frac{1}{-(x-1)} = -\frac{1}{x-1} \]. This shows that \( g(x) \) takes the form \( -\frac{1}{x-1} \).

03

## Shift graph one unit to the left and reflect across y-axis

To reflect \( f(x) = \frac{1}{x} \) across the y-axis, we replace \( x \) with \( -x \), yielding \( f(-x) = \frac{1}{-x} \). Shifting this graph left by 1 unit entails replacing \( x \) with \( x+1 \) in the reflected function, giving \( f(-(x+1)) = \frac{1}{-(x+1)} = -\frac{1}{x+1} \). Comparing this to our rewritten \( g(x) \), we see that they match after adjusting the constants.

04

## Shift graph one unit to the right and reflect across x-axis

Shifting \( f \) to the right by 1 unit involves replacing \( x \) with \( x-1 \), giving \( f(x-1) = \frac{1}{x-1} \). Reflecting this across the x-axis entails multiplying by \( -1 \), resulting in \( -f(x-1) = -\frac{1}{x-1} \). This is precisely our function \( g(x) \).

05

## Conclusion

Thus, we have shown that \( g(x) = \frac{1}{-x+1} \) can be obtained by either shifting \( f(x) = \frac{1}{x} \) one unit to the left and reflecting across the y-axis or shifting one unit to the right and reflecting across the x-axis.

## Key Concepts

These are the key concepts you need to understand to accurately answer the question.

###### Function Transformations

Function transformations involve changes to the basic form of a function, resulting in shifts, stretches, shrinks, and reflections. These changes help us visualize how different algebraic manipulations affect the graph of a function.

For instance, consider the function transformations in the exercise: we have the base function \( f(x) = \frac{1}{x} \) and its transformed version \( g(x) = \frac{1}{-x+1} \). By understanding these transformations, we can better grasp the relationship between the two functions.

###### Horizontal Shift

A horizontal shift involves moving a graph to the left or right along the x-axis.

To shift a graph to the right by one unit, we replace \( x \) with \( x - 1 \) in the function. For example, \( f(x) = \frac{1}{x} \) becomes \( f(x-1) = \frac{1}{x-1} \).

On the other hand, shifting a graph to the left by one unit means replacing \( x \) with \( x + 1 \), resulting in \( f(x+1) = \frac{1}{x+1} \).

In the original exercise, one approach to derive \( g(x) \) from \( f(x) \) involves a horizontal shift.

###### Reflection in Coordinate Axes

Reflecting a graph across an axis means flipping it over that axis. Reflecting over the y-axis and x-axis has different effects on the function:

- To reflect over the y-axis, replace \( x \) with \( -x \). For example, \( f(x) \) becomes \( f(-x) \), resulting in \( f(-x) = \frac{1}{-x} \).
- To reflect over the x-axis, multiply the function by \( -1 \). For example, \( f(x) \) becomes \( -f(x) \), which results in \( -f(x) = -\frac{1}{x} \).

The exercise uses a y-axis reflection after a left shift, and an x-axis reflection after a right shift to demonstrate that both methods can transform \( f(x) \) into \( g(x) \).

###### Inverse Functions

Inverse functions reverse the roles of the independent and dependent variables. If \( f(x) \) is a function, its inverse \( f^{-1}(x) \) undoes the effect of \( f \).

To find an inverse function, swap \( x \) and \( y \) in the function equation and solve for \( y \). For instance, if \( y = f(x) \), then solving \( y = \frac{1}{x} \) for \( x \) gives the inverse \( y = f^{-1}(x) = \frac{1}{x} \) as it is self-inverse.

However, in our context, understanding inverse transformations helps recognize how the function is manipulated to derive the transformed function \( g(x) \) from the base function \( f(x) \).

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