Using the Mean Value Theorem

Question

1.

x: 1 3 5 7 9

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f(x): 0 6 18 29 42

Selected values of a differentiable function ƒ are given in the table above. What is the fewest possible number of values of c in the interval [1,9] for which the Mean Value Theorem guarantees that ƒ′(c)=6?

a) zero

b) one

c) two

d) three

2.

The Mean Value Theorem can be applied to which of the following functions on the closed interval [−5,5]?

a) ƒ(x) = 1/sinx

b) ƒ(x) = x-1/|x-1|

c) ƒ(x) = x²/x²-36

d) ƒ(x) = x²/x²-4

Answer 1

f is differentiable on the interval [3,9]. So, by the mean value theorem there is at least one value c in the interval (3,9) such that f'(c) = [f(9) - f(3)] / (9-3) = (42-6) / 6 = 6..So, the answer is b.

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f(x) = x² / (x² - 36) is the only one of the four functions that is differentiable on the interval [-5,5].