First slide

Differentiability

Question

$If f (\frac{x + y}{2}) = \frac{f (x) + f (y)}{2} for all real x and y and f^{1} (0) exists and equals to - 1 and f (0) = 1,$ then $f^{1}$ (x)=

Moderate

A

1
B

0
C

-1
D

2

Solution

$f (\frac{x + y}{2}) = \frac{f (x) + f (y)}{2} .... (1)$
Put y=0, f(0)=1
$f (\frac{x}{2}) = \frac{1}{2} (f (x) + 1) \Rightarrow f (x) = 2 f (\frac{x}{2}) - 1 \to (2)$
$f' (x) = \underset{h \to 0}{l t} \frac{f (x + h) - f (x)}{h}$
$= \underset{h \to 0}{l t} \frac{f (\frac{2 x + 2 h}{2}) - f (x)}{h}$
$= \underset{h \to 0}{l t} \frac{\frac{f (2 x) + f (2 h)}{2} - f (x)}{h}$
$= \underset{h \to 0}{l t} \frac{f (2 x) + f (2 h) - 2 f (x)}{2 h}$
$= \underset{h \to 0}{l t} \frac{2 f (x) - 1 + f (2 h) - 2 f (x)}{2 h}$
$= \underset{h \to 0}{l t} \frac{f (2 h) - 1}{2 h} = f' (0)$
$f' (x) = - 1$

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