If $f(x_0)+f'(x_0)(x-x_0)$ is the affine approximation to $f$ at $x_0$, is $f'(x_0)(x-x_0)$ the linear approximation?

Question

If $f: \mathbb R \to \mathbb R$ is differentiable at $x_0$, the affine approximation of $f$ at $x_0$ is $f(x_0)+f'(x_0)(x-x_0)$.

So, would that mean that $f'(x_0)(x-x_0)$ is the linear approximation?

Answer 1

I would say only if $f(x_0) = 0$, if you mean linear in the sense of linear algebra, in contrast to affine.

$T(x) = f(x_0) + f'(x_0)(x-x_0)$ is the best first order approximation, this is also called linear. Also functions $y = m x + n$ are called linear, the special case $y = m x$ proportional.

The graph of $T$ is a line, which for $f(x_0) \ne 0$ is not containing the origin.

In the terms of linear algebra this is not linear, but affine. And these two different kinds of "linear" seem to be the problem here.

Answer 2

If $f$ has a derivative near $x_0$ then it can be approximated by an affine function but not necessarily linear(i.e. crosses zero at $x_0$). Sometimes you may find people referring to affine as linear but don't bother with the terms remember that it has the derivative as a slope and it crosses the point near which it was defined.

Answer 3

The expression $f'(x_0) \, (x-x_0)$ is a linear approximation of the difference $f(x)-f(x_0)$.

This is the same thing as an affine approximation of $f(x)$, so people don't bother very much with the distinction.