- #1
transgalactic
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if f(x) is differentiable on x_0
prove that
http://img102.imageshack.us/img102/3189/51290270sj1.th.gif
??
[itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} [/itex]
now let [itex] u = x+h \Rightarrow x = u-h [/itex]
So [itex] f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h} [/itex]
Since [itex] h \to 0, f'(u-h) \to f'(u) [/itex]
Therefore [itex] f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h} [/itex]
[itex] f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)[/itex]
[itex] 2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h} [/itex]
Hence, [itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h} [/itex]
i understand this solution but
how to pick those variables??
and why it didnt use the point x_0
that they presented
prove that
http://img102.imageshack.us/img102/3189/51290270sj1.th.gif
??
[itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} [/itex]
now let [itex] u = x+h \Rightarrow x = u-h [/itex]
So [itex] f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h} [/itex]
Since [itex] h \to 0, f'(u-h) \to f'(u) [/itex]
Therefore [itex] f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h} [/itex]
[itex] f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)[/itex]
[itex] 2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h} [/itex]
Hence, [itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h} [/itex]
i understand this solution but
how to pick those variables??
and why it didnt use the point x_0
that they presented
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