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Chapter 2: Problem 52

Use transformations of \(f(x)=\frac{1}{x}\) or \(f(x)=\frac{1}{x^{2}}\) to graph each rational function. $$g(x)=\frac{1}{(x+1)^{2}}$$

Short Answer

Expert verified
The graph of the given function \(g(x)= \frac{1}{(x+1)^{2}}\) will be a U-shaped figure open upwards, crossing the y-axis at point (0,1) with its lowest point at (-1,1), obtained by shifting the graph of the base function \(f(x)= \frac{1}{x^{2}}\) one unit to the left.

Step by step solution

01

Identify the Base Function

The base function in this exercise is \(f(x)= \frac{1}{x^{2}}\). The graph of this function is a U-shape open upwards, crossing the y-axis at 1.
02

Recognizing the Transformation to be done

In the function \(g(x)= \frac{1}{(x+1)^{2}}\), compared to the base function \(f(x)= \frac{1}{x^{2}}\), the 'x' in the denominator is replaced with '(x+1)'. This represents a shift of the graph of base function one unit to the left.
03

Graphing the Transformed Function

To graph \(g(x)\), take the graph of \(f(x)= \frac{1}{x^{2}}\) and shift it one unit to the left. The transformed graph will still be a U-shape open upwards like that of base function but now it will cross the y-axis at point (0,1) and have its lowest point at (-1,1).

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Most popular questions from this chapter

Galileo's telescope brought about revolutionary changes in astronomy. A comparable leap in our ability to observe the universe took place as a result of the Hubble Space Telescope. The space telescope was able to see stars and galaxies whose brightness is \(\frac{1}{50}\) of the faintest objects observable using ground-based telescopes. Use the fact that the brightness of a point source, such as a star, varies inversely as the square of its distance from an observer to show that the space telescope was able to see about seven times farther than a groundbased telescope.

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