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Verify the Mean Value Theorem for f(x) = x³ - 2x on [0, 2]…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Verify the Mean Value Theorem for $f(x) = x^3 - 2x$ on $[0, 2]$ . Find the slope of the chord and the point $c$ in $(0, 2)$ where the tangent has that slope, to four decimal places.

Need a reminder? The key ideaShow

Mean Value Theorem: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then $f'(c) = \dfrac{f(b) - f(a)}{b - a}$ for some $c$ in $(a, b)$ . It gives bounds: if $m \le f'(x) \le M$ then $m(b - a) \le f(b) - f(a) \le M(b - a)$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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