Geometric Interpretation of a "Near"-MVT Going through Larson's Problem Solving Through Problems, I am asked to give a geometric interpretation of the result below.  I have been sketching it, and only got so far as to note that there must be a local minimum or maximum at $c$, although I cannot get very much further then that. 
Let $f$ be differentiable with $f'$ continuous on $[a,b]$.  
If $\exists \text{ } c\in(a,b]$ such that $f'(c)=0$, then we can find $d\in(a,b)$ such that $$f'(d)= \frac{f(d)-f(a)}{b-a}$$
 A: I picked up a copy of Larson's book and found exercise 6.6.4.
I found no task to give a geometrical interpretation.
Rather Larson looks at
$$
F(x) = f'(x) - \frac{f(x) - f(a)}{b - a}
$$
Then a $d \in (a, b)$ with $F(d) = 0$ would fulfill the given condition.
He then distincts three cases, depending on how $f(a)$ compares to $f(c)$ and tries to find points 
$x_1, x_2 \in (a,b)$ with $F(x_1) > 0$ and $F(x_2) < 0$ for 
application of the MVT.
A: 
This should be clear enough. If you take a look, $f'(d)=\frac{f(b)-f(a)}{b-a}$.
A: Let $g(x)=e^{-x/(b-a)}(f(x)-f(a))$. 
$g'(d)=0$ is equivalent to  $f'(d) - \frac{f(d)-f(a)}{b-a} = 0$ by direct calculation.  The question must be whether an interior extremum exists in the interval, at which $g'(d)=0$.  The only way that can be avoided is if $g$ is strictly increasing or decreasing: 
$f(x) - f(a) = H(x) e^{x/(b-a)}$ with $H$ increasing (decreasing).
Maybe there is a contradiction between having a point with $f'(c)=0$ and $H$ strictly monotone but I don't see it at the moment. $H$ could be positive and decrease rapidly near $a$, causing negative derivative of the product near $a$, then slow down to a near-constant so that the exponential factor forces an increase near $b$, which by Darboux's theorem means a point with $f'(c)=0$ inside the interval. 
If I have not made a mistake it looks like Larson's statement is incorrect.
