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What type of symmetry does the function $y=\frac{1}{|x|}$ have? Specify the intervals over which the function is increasing and the intervals where it is decreasing.

user35603
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Aly
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  • Hi, and welcome to M.SE. You'll get more appropriate help if you include what you've tried, and what you're struggling with. For example, do you know the definition of even / odd, or increasing / decreasing? Also, please note that many will find the use of the imperative to be rude; please consider editing your question. –  Sep 02 '13 at 02:00
  • Sorry about that! – Aly Sep 02 '13 at 02:05

2 Answers2

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Let $f(x)=y$

HINT: what does $f(-x)$ equal?

Let's look at the function, $f(x)=\frac{1}{x}$. Now, a function is even if: $$f(-x)=f(x)$$ and a function is odd if: $$f(-x)=-f(x)$$ We therefore can see that this function is an odd function, since; $$f(-x)=\frac{1}{-x}=-\frac{1}{x}=-f(x)$$ Now, the absolute value is defined as $$|x| = \sqrt{x^2} $$ With this in mind, let's look at your function; $$f(x)=\frac{1}{|x|}$$ Now what happens when we look at $f(-x)$? $$f(-x)=\frac{1}{|-x|}=\frac{1}{\sqrt{(-x)^2}}=\frac{1}{\sqrt{x^2}}=\frac{1}{|x|}=f(x)$$ Thus, $f(x)=\frac{1}{|x|}$ is an even function, symmetric about the y-axis.

Eleven-Eleven
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Plotting the function is a good way to guess at where the function is going to be increasing or decreasing. Then you can think about why.

As you increase $x$, what is going to happen to $\frac{1}{|x|}$? What happens when $x$ approaches (but does not equal!) 0?

tylerc0816
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