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Could you show me an easy and fast way to test whether a function is odd or even using the calculator ?

I've only came across the graphing way and another algebraic way and I'm afraid to make a silly mistake while doing them.

N. F. Taussig
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carlos
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3 Answers3

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Great question.

I myself had some difficulty understanding the difference between even and odd functions. I found this PurpleMath article on even/odd functions pivotal in my understanding of them.

For finding even/odd function on your calculator, I just used the function manager of a TI-84.

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The function shown is from the first example on that PurpleMath page I linked, -3x^2+4.

I then go to the table using 2nd/GRAPH, and input a value of x and -x as independent values.

enter image description here

As you can see, f(x) and f(-x) are both -296, thus the function is even, as the PurpleMath article corroborates.

You may be wondering: "Well, what about odd and neither? How do I determine those?" As stated in the PurpleMath article, in an odd function f(-x) will be the exact opposite of what we started with, or f(x). In a function that's neither even nor odd, f(-x) will not be the same or the opposite of f(x).

Zulfe
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Does the definition count?

Odd functions are functions such that $$f(-x) = -f(x)$$ and even functions are such that $$f(-x) = f(x)$$ for all $x$ in the domain of $f$.

Graphically, odd functions are symmetric about the origin and even functions are symmetric about the $y$-axis.

Em.
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  • Hi i know this but the problem is i don't really know how to graph some functions on the axis and i don't want do a silly mistake when i determine them using algebra(factoring and simplified and etc) ,, so is there any way to know the type of any function quickly using a calculator ? thanks – carlos Mar 10 '16 at 02:38
  • The short answer is no. There might be, but it wouldn't be "easy". It would still take work. Also, I think it has to be an arbitrary $x$. For example, let $f(x) = x^2$ and let $t$ be some arbitrary element in the domain of $f$. Then $f(t) = t^2$ and $f(-t) = (-t)^2 = t^2 = f(t)$ and hence, $f$ is even. – Em. Mar 10 '16 at 02:47
  • Oh, sorry I guess I misread your question. I promise I am literate. Hahaha. Using a calculator, check that the graph satisfies the symmetries I have highlighted. Sorry, haha. – Em. Mar 10 '16 at 02:52
  • thanks but my calculator doesn't graph though and i have an exam coming soon :( – carlos Mar 10 '16 at 02:56
  • Oh!! You should explain that your calculator is not a graphing calculator. I see. Then 1. Do what I said three comments about. 2. Show that a function is not even or odd by finding a particular value that doesn't work. Non-rigorously, plug in several values and use the calculator to see if it is even or odd using the definition above. I do not recommend this. – Em. Mar 10 '16 at 03:00
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Try the following: Pick a random value $x$ such that $x$ and $-x$ are in your domain, which I assume here is all of $\mathbb{R}$. Use your calculator to evaluate $f(x)$ and $f(-x)$.

If $f(x)=f(-x)$ your function is even. If $f(x)=-f(-x)$ your function is odd. If neither of these relations hold, then neither is true.

WARNING: Just because one of these relations holds for a specific choice of $x$, does not mean it holds for all $x$. This is a good way of checking your work, but the algebraic way is generally the only full way to check this.

Alekos Robotis
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