Question

Simplifique [F(x+h) - F(x)] / h sendo F(x) igual a:
F(x) = 1/x²
F(x) = 1/x+2

Answer 1

$\boxed{\boxed{ \frac{f(x+h)-f(x)}{h} }}$


a)  lembrando algumas propriedades : 
$\bmatrix (a\pm b)^2=a^2\pm2ab+b^2\\\\(a^2b^2)= (ab)^2\\\\ a^2-b^2=(a+b)(a-b)\\\\ \frac{a}{b}\pm \frac{c}{d}= \frac{ad\pm cb}{bd} \end$


f(x+h) -> substitui o x da função por x+h

temos:
$\frac{ \frac{1}{(x+h)^2} - \frac{1}{x^2} }{h} \\\\ = \left( \frac{1}{(x+h)^2} - \frac{1}{x^2}\right) \frac{1}{h} \\\\ = \left( \frac{x^2-(x+h)^2}{(x+h)^2x^2}\right) \frac{1}{h} \\\\ = \left( \frac{[x+(x+h)] [(x-(x+h)]}{(x+h)^2x^2}\right) \frac{1}{h} \\\\ = \left( \frac{(2x+h)(-h)}{(x+h)^2x^2}\right) \frac{1}{h} \\\\ = \left( \frac{(2x+h)(-\not h)}{(x+h)^2x^2}\right) \frac{1}{\not h} \\\\ = \frac{(2x+h)(-1)}{(x+h)^2x^2} \\\\ = \frac{-(2x+h)}{[x(x+h)]^2}$


b)

$f(x)= \frac{1}{(x+2)} \\\\:::::::::::::::::::::::::::::: \\\\ \frac{ \frac{1}{x+h+2}- \frac{1}{x+2} }{h} =\\\\ = \left( \frac{1}{x+h+2} - \frac{1}{x+2} \right) \frac{1}{h} \\\\ = \left( \frac{(x+2)-(x+h+2)}{(x+h+2)(x+2)} \right) \frac{1}{h} \\\\ = \left( \frac{x+2-x-h-2}{(x+h+2)(x+2)} \right) \frac{1}{h} \\\\ = \left( \frac{- \not h}{(x+h+2)(x+2)} \right) \frac{1}{\not h} \\\\ = \frac{-1}{(x+2)(x+h+2)}$