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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

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    "... 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).2011-12-06

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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.

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    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.2011-12-05
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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.

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    If a function is linear then it is affine(convex).2011-12-06
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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.

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    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.2011-12-06
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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.