The Mean Value Theorem for $f\in \mathbb R ^{\mathbb R}$, is stated only for $a < b$ as:
If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists $x\in (a,b)$ such as: $$f(b)-f(a) = f'(x) (b-a)$$
As far as I know, it cannot be stated if $b< a$, or does it ? I mean for example if we take $a$ and $b$ in a set $[A,B]$ we can't state the Mean Value Theorem for any $a$ and any $b$ in $[A,B]$ ?
However, I suspect and am pretty sure, that if $b<a$ the following version of the theorem is true:
$$|f(b)- f(a)| = |f'(x)| |b-a|$$
However, I cannot convince myself that this adapted version is true.
Here are my thoughts on this:
- First, does moving $b$ to the left of $a$ mean we are doing, like if there were two cursors (one behind $(a,f(a))$ another behind $(b,f(b))$), a switch between the two cursors ? In other words, is it like an axial symmetry (vertical axis, passing through $(a+b)/2$) performed on the graph ?
- second, my father has been telling me that I should consider the four cases (depending on the sign of $(b-a)$ and of $(f(b)-f(a))$) and see that the absolute value version of MVT hold in all the cases.
Thanks a lot in advance.
If $b<a$, consider the interval $(b,a)$ and use the mean value theorem to write $f(a)-f(b)=f'(x)(a-b)$ for some $x\in (b,a)$. This gives us $-(f(b)-f(a))=f'(x)(-(b-a))$ and so $f(b)-f(a)=f'(x)(b-a)$, so the theorem can go unchanged as long as we specify the correct interval $(b,a)$.
One way to think of this is that slopes and derivatives allow us to track directional change (or at least estimate directional change near a point). In the linear equation $f(x)=mx$, if I move forward from $a$ to $b$ by the amount $(b-a)$, this change in the $x-$direction is scaled by the slope $m$ and we get the corresponding change $f(b)-f(a)=m(b-a)$ in the $y-$direction. However, if we move backwards from $b$ to $a$ by the amount $a-b$, the slope still allows us to track this change in the $y-$direction as $f(a)-f(b)=m(a-b)$. The slope works generally for all directional movement (forwards or backwards in this case) to track the corresponding change to the output of a function.
The derivative works in an analogous way, since it gives us the slope of a line of nearby approximation to the graph (the tangent line). Our scaling value/slope for our approximation at $x$ is $f'(x)$, and it works to estimate changes in output in both the forward and backward direction on the $x-$axis, just like the slope of a linear function does.
The MVT relates the total rate of change over an interval (slope of the secant line between two points of the graph) to the locally approximated rate of change of a point between them (the slope of some tangent line in the middle, i.e. the derivate at that point). Because each of these values is flexible enough to describe both forwards and backwards movement, the theorem holds for both forwards movement from $a$ to $b$ with $a<b$ and backwards movement from $a$ to $b$ with $b<a$.
Hope this helps :)