Subsection 2.13.3 (Optional) — Why is the MVT True
We won't give a real proof for this theorem, but we'll look at a picture which shows why it is true. Here is the picture. It contains a sketch of the graph of \(f(x)\text{,}\) with \(x\) running from \(a\) to \(b\text{,}\) as well as a line segment which is the secant of the graph from the point \(\big(a\,,f(a)\big)\) to the point \(\big(b\,,f(b)\big)\text{.}\) The slope of the secant is exactly \(\frac{f(b)-f(a)}{b-a}\text{.}\)
Remember that we are looking for a point, \(\big(c\,,f(c)\big)\text{,}\) on the graph of \(f(x)\) with the property that \(f'(c)=\frac{f(b)-f(a)}{b-a}\text{,}\) i.e. with the property that the slope of the tangent line at \(\big(c\,,f(c)\big)\) is the same as the slope of the secant. So imagine that you start moving the secant upward, carefully keeping the moved line segment parallel to the secant. So the slope of the moved line segment is always exactly \(\frac{f(b)-f(a)}{b-a}\text{.}\) When we first start moving the line segment it is not tangent to the curve — it crosses the curve. This is illustrated in the figure by the second line segment from the bottom. If we move the line segment too far it does not touch the curve at all. This is illustrated in the figure by the top segment. But if we stop moving the line segment just before it stops intersecting the curve at all, we get exactly the tangent line to the curve at the point on the curve that is farthest from the secant. This tangent line has exactly the desired slope. This is illustrated in the figure by the third line segment from the bottom.