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f (x)=x^2-2х
f’= (f(x+h) - f(x))/h

Sagot :

LammettHash

If you're asking about the difference quotient alone, you have

f (x) = x ² - 2x

which means

f (x + h) = (x + h)² - 2 (x + h)

so that

(f (x + h) - f (x)) / h = (((x + h)² - 2 (x + h)) - (x ² - 2x)) / h

… = (x ² + 2xh + h ² - 2x - 2h - x ² + 2x) / h

… = (2xh + h ² - 2h) / h

*and you would stop here*.

On the other hand, if you're talking about the derivative (which I believe you are, since your question says f '), take the limit as h approaches 0. In particular, this means h ≠ 0, so there is some cancellation:

(f (x + h) - f (x)) / h = (2xh + h ² - 2h) / h

… = 2x + h - 2

Then as h approaches 0, you're left with

f ' (x) = 2x - 2

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