Quod Erat Demonstrandum

2009/03/30

Simple questions about mean value theorem

Filed under: HKALE,Pure Mathematics — johnmayhk @ 5:20 下午
Tags: , ,

For your revision, students.

Question 1

Suppose f(1) = f(2) = 0, f(3) = 1 and f is twice differentiable on [0,3].

Show that f''(c) > \frac{1}{2}

for some c \in (0,3).

Question 2

Suppose f(0) = 0, f(1) = 1, f is differentiable on [0,1].

Show that \frac{1}{f'(a)} + \frac{1}{f'(b)} = 2

for some a, b \in (0,1).

Solution to question 1

0 = f(2) - f(1) = f'(a) for some a \in (1,2)
1 = f(3) - f(2) = f'(b) for some b \in (2,3)

Hence

1 = f'(b) - f'(a) = f''(c)(b - a) < f''(c)(3 - 1) for some c \in (a,b)
\Rightarrow \frac{1}{2} < f''(c) for some c \in (0,3).

Solution to question 2

\because f(0) = 0, f(1) = 1 and f is continuous on [0,1],

f(c) = \frac{1}{2} for some c \in (0,1).

Now

f(c) - f(0) = f'(a)c for some a \in (0,c)
f(1) - f(c) = f'(b)(1 - c) for some b \in (c,1)

Thus,

\frac{1}{f'(a)} + \frac{1}{f'(b)} = \frac{c}{f(c) - f(0)} + \frac{1 - c}{f(1) - f(c)} = 2

for some a, b \in (0,1).

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