Determine whether the "Mean Value Theorem" applies to f(x) = - 1/x on the interval [-3, -1/2] .?

Question

If the "Mean Value Theorem" applies,

find all value(s) of c in the interval such that

f'(c) = [ f(-1/2) - f(-3)] / [ (-1/2) - (-3) ].

If the "Mean Value Theorem" does not apply, state why.

Answer 1

The MVT *always* applies to *any* function that is differentiable on the closed interval [a,b].

f(x) = -x^(-1) ==> f'(x) = x^(-2)

(x(-1/2) - f(-3))/((-1/2) - (-3)) = ((2 - (1/3))/(3 - (1/2)) = .66666

x^(-2) = .66666 ==> x² = (3/2) ==> x = ±1.2247 so
x = -1.2247 is the point at which f'(x) = avg slope.


Doug

Answer 2

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Answer 3

Doug's answer is mostly correct except that it also applies on the interval [a,b). In other words, it need not be differentiable at b. Your math professor will not agree with this.