Example 1: Suppose you have the function f(x) = x^2 , and you’re looking at the interval [1, 3]. The Mean
Value Theorem for Derivatives (this is often abbreviated MVTD) states that there is some number c such
that
f′(c) = = 4Because f ́(x) = 2x, plug in c for x and solve: 2c = 4 so c = 2. Notice that 2 is in the interval. This is what
the MVTD predicted! If you don’t get a value for c within the interval, something went wrong; either the
function is not continuous and differentiable in the required interval, or you made a mistake.
Example 2: Consider the function f(x) = x^3 − 12x on the interval [−2, 2]. The MVTD states that there is a
c such that
f′ (c)= = −8Then f′(c) = 3c^2 − 12 = −8 and c = ± (which is approximately ±1.155).
Notice that here there are two values of c that satisfy the MVTD. That’s allowed. In fact, there can be
infinitely many values, depending on the function.
Example 3: Consider the function f(x) = on the interval [−2, 2].
Follow the MVTD.