Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm inclined to say yes, as it doesn't involve exponentiation, roots, logarithmic or trigonometric functions, but I watched a video where the teacher said that the absolute value function is "clearly non-linear". Why would he say that? Is he wrong?

Wikipedia's graph for abs:

enter image description here

share|improve this question
Well what's your definition of linear function? –  The Chaz 2.0 Dec 5 '11 at 21:43
I'd say a function that draws a straight line when graphed on the two/three dimensional plane. –  omgzor Dec 5 '11 at 21:48
Well abs(x) doesn't satisfy that! –  The Chaz 2.0 Dec 5 '11 at 21:49
Two things:that article doesn't show (just) one straight line! Also, a graphical definition usually isn't best. How about defining a linear function as a first degree polynomial? In that case, we can see that no first degree polynomial satisfies the properties of abs(x). One such property is that abs(x) is an even function, whereas f(x) = ax + b is neither even nor odd (with $a \neq 0 \neq b$) –  The Chaz 2.0 Dec 5 '11 at 21:58
"... as it doesn't involve exponentiation, roots, logarithmic or trigonometric functions." We have to be very careful when defining things in terms of what they are not. For example, $y=\frac{1}{1+x^2}$ doesn't involve anything in your list, but that's not linear (graph it!). However, $y = \ln (e^{3x})$ turns out to be linear (simplify it to see what linear function it's equivalent to). –  Shaun Ault Dec 6 '11 at 0:36

5 Answers 5

up vote 6 down vote accepted

Linear functions in analytic geometry are functions of the form $f(x)=a\cdot x+b$ for $a,b \in \mathbb{R}$.

Now try to write $\text{abs}(x)$ in such a form.

Another way to see it: linear functions are "straight lines" in the coordinate system (excluding vertical lines), this clearly excludes having a "sharp edge" in the graph of the function like $\text{abs}(x)$ has it for $x=0$.

In linear algebra (and this is the more common definition) linear functions denote ones of the form $f(x)=a\cdot x$ which is equivalent to require $b=0$ in the above definition. As $\text{abs}(x)$ is not linear with the first, weaker definition it cannot be linear either with this definition.

share|improve this answer
This is usually called an affine function; it is linear only if $b = 0$. –  student Dec 5 '11 at 21:43
I did google, and read that same link. Thanks for being insulting. –  omgzor Dec 5 '11 at 21:44
@Leandro Yea that is also in the article. But if $\text{abs}$ is not even affine it can neither be linear (using your definition) so it is good to use the weaker one for the question he asked. –  Listing Dec 5 '11 at 21:45
@omgzor I did not mean to be insulting, if you read that link why don't you include your efforts into your question and tell us which part you don't understand? –  Listing Dec 5 '11 at 21:46
It depends a great deal what level you are working at. The equation of a general straight line in the plane has the form $ax+by=c$ (deals with the vertical case) or some equivalent. In three dimensions you need to intersect two planes etc. Given that this is for enquiries at every level, I would note that my daughter is learning to distinguish between linear and quadratic equations (or forms to generalise at a higher level). OP does not deal with definition or level, so it is easy both to patronise and also to be over-sophisticated. –  Mark Bennet Dec 5 '11 at 22:24

A function $f(x)$ is linear if it satisfies the property $$f(ax+y) = af(x) + f(y).$$ Let's try $a=-1$, $x=1$, $y=0$: $$\begin{align*} |ax+y| = |-1| &= 1\\ -1\ |1|+|0| &= -1\end{align*},$$ so $f(x) = |x|$ is not linear.

Sometimes (especially in geometry) "linear" is understood to mean affine. A function $f(x)$ is affine if it satisfies the property $$f[ax + (1-a)y] = af(x) + (1-a)f(y).$$ Once again let's try $a=-1$, $x=1$, $y=0$: $$\begin{align*} |ax+(1-a)y| = |-1| &= 1\\ af(x)+(1-a)f(y) = -1\ |1| + 2\cdot 0 &= -1,\end{align*}$$ so $|x|$ isn't affine either.

share|improve this answer
It is worse than failing $f(ax+y) = af(x) + f(y)$. It fails BOTH $f(x+y) = f(x) + f(y)$ and $f(ax) = af(x)$. –  GEdgar Dec 5 '11 at 22:46
If a function is linear then it is affine(convex). –  Hassan Muhammad Dec 6 '11 at 7:07

I would simply define a linear function as always having the same slope (and, more technically, it must be continuous).

Clearly, the absolute value function has a negative slope for values < 0 and positive slope for values > 0. So it's not linear.

share|improve this answer
In order for a function to have a well-defined sense of "slope", it must be differentiable, which implies continuity. However, the absolute value function, in addition to not having the same slope everywhere, does not even have a slope at 0, so your idea still works. –  Calvin McPhail-Snyder Dec 6 '11 at 0:32

Think of the definition of absolute value. It is a piecewise defined function. |x| = x, if x>=0 and -x if x<0. In other words, the graph of y=|x| is formed by two pieces of two lines. For the part of the domain where x-values are less than zero, the graph corresponds to the graph of y=-x. For parts of the domain where x-values are greater than or equal to zero, the graph corresponds to the graph of y=x. While the absolute value function does not satisfy the above definitions for linear functions, it is actually "parts" of two linear functions.

share|improve this answer

For people that don't understand everything being said, simply if you draw a horizontal line on the graph and it does not cross the plotted point more than once you have a linear function.

share|improve this answer
Not true at all (if I'm understanding you correctly). $f(x) = x^3$ is one-to-one.. Also why would you dig up such an old post that has an accepted answer? –  Cameron Williams Nov 27 '13 at 2:49
This only works when referring to single parabolas and absolute value functions not a double parabola. –  user111737 Nov 27 '13 at 2:51
The main problem is that you phrased your answer as a "sufficient condition", when it is in fact a "necessary condition" (for nonconstant functions [thanks! @TrevorWilson]). The statement "for a nonconstant linear function, if you draw a horizontal line on the graph it must touch exactly one point" is true, and is in fact applicable to the present case. What you wrote, however, is the reverse implication, which is false. –  Willie Wong Nov 27 '13 at 8:48
@WillieWong It's not even a necessary condition, because the constant zero function is linear. –  Trevor Wilson Nov 28 '13 at 19:37
@TrevorWilson indeed you are right. I will cheat with my mod powers to correct the statement above. –  Willie Wong Nov 29 '13 at 8:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.