Lagrange’s Mean Value Theorem

Question

From MVT:

$f'(c) = \frac{f(b)-f(a)}{b-a}$.

Given function

$y=(x+1)(x-1)^2\text{ ; }x\in[-1,a]; a\geq0$

find out how the number of $c$ is changing depending on $a$.

For example we know that for $a=\frac{1}{2}$, just one value $c$ exists, for $a=2$, there are two values $c$. How can I find out where the breakpoints are?

Answer 1

Hint: I think the point with the exercise is that you draw the graph and do a graphical interpretation of MVT.

1. Draw the graph.
2. Start moving $a$ to the right from $-1$ and interpret the RHS as a slope of the line from $(-1,f(-1))$ to $(a,f(a))$.
3. For each $a$, figure out if you can move the line parallel to make it tangent at some other point(s) between $-1$ and $a$. Those points are possible $c$.

Once you do it, the breakpoints become obvious.