# Quod Erat Demonstrandum

## 2009/03/30

### Simple questions about mean value theorem

Filed under: HKALE,Pure Mathematics — johnmayhk @ 5:20 下午
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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 $a, b \in (0,1)$.