Using the principle of the mean-value theorem on the indicated interval, how do you find all numbers c that satisfy the conclusion of the theorem $f (x) = x^{3} - 2 x^{2}$ $f(x)=x^3-2x^2$ ; [0, 2]?

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Answer 1 · 2016-10-13T14:04:37

The conclusion of the Mean Value Theorem for $f$ $f$ on $[a, b]$ $[a,b]$ says:

there is a $c$ $c$ in $(a, b)$ $(a,b)$ such that $f'(c) = (f(b)-f(a))/(b-a)$

(The theorem make no guarantees about our ability to find the value(s) of $c$ )

Find the value(s) of $c$ does not use the Mean Value Theorem, it uses the derivative and some algebra.

For $f(x) = x^3-2x^2$ on $[0,2]$ , the conclusion of MVT says:

there is a $c$ in $(0,2)$ such that $f'(c) = (f(2)-f(0))/(2-0)$

To find the $c$ (or $c$ 's) find $f'(x)$ , do the arithmetic on the right and solve the resulting equation. In this case, solve

$3x^2-4x = 0$

The solutions are $0$ and $4/3$ .

The $c$ mentioned in MVT must be in $(0,2)$ , so $4/3$ is the only value of $c$

Using the principle of the mean-value theorem on the indicated interval, how do you find all numbers c that satisfy the conclusion of the theorem f(x)=x^3-2x^2f(x)=x3−2x2f(x)=x^3-2x^2; [0, 2]?