# How can I generalize a picture for the Mean Value Theorem to the Generalized MVT?

How can I transmogrify this figure for the Generalized MVT? If $f$ and $g$ are continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then $\exists$ $c ∈ (a, b) \ni \cfrac{f'(c)}{g'(c)} = \cfrac{f(b)-f(a)}{g(b)-g(a)}$.

Please answer with a picture. Please edit my scan, and draw or write in what you mean. Don't just change the t-axis to $g(t)$.

*Calculus: The Language Of Change* (2005)
by David W. Cohen, James M. Henle. pp. 827-829. The original colored in just blue. I annotated and added more colors.

## 1 answer

Plotting $f$ and $g$ together is unlikely to lead to intuition about the GMVT; they're just two arbitrary curves, whereas the value from that diagram of the MVT comes from the fact that the yellow line is constructed to give intuition. The art of drawing intuitive figures is largely about figuring out what you need to construct in addition to the raw givens. So I'm going to suggest, not a ‘transmogrification’ of that particular construction, but a different construction that targets the general case.

Consider $f_1(x) = f(x) - k \cdot g(x)$, where $k = \frac{f(b) - f(a)}{g(b) - g(a)}$. If $f_1'(c) = 0$, then $f'(c) = k \cdot g'(c)$, and so $c$ satisfies the GMVT condition. It's easy to see that $f_1(a) = f_1(b)$, so an illustration of looking for some $c$ where $f_1'(c) = 0$ on a plot of $f_1$ over $[a,b]$ is just the figure for Rolle's theorem, as it appears in that large screenshot.

(P.S. Please avoid making screenshots the bulk of your post in future.)

#### 2 comments

Thanks. Can you please edit my scan and draw or write in what you mean?

The graph in the upper-right corner of the proof box is all you need. Imagine that's a graph of $f_1$. Once you have $f_1$, the details of $f$ and $g$ don't matter and graphing them is neither necessary nor useful, though you can always imagine what that would look like if you want.

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