Breabel
contestada

Let f(X)=1/x+1

Use the limit definition of the derivative to find:

i) f(-4)
ii) f(-3)
iii) f(1)
iv) f(3)

Respuesta :

[tex]\bf f(x)=\cfrac{1}{x+1}\qquad \qquad \stackrel{d e f in i tion~of~a~derivative}{\lim\limits_{h\to 0}~\cfrac{f(t+h)-f(t)}{h}} \\\\\\ \lim\limits_{h\to 0}~\cfrac{\frac{1}{x+h+1}-\frac{1}{x+1}}{h}\implies \cfrac{\frac{(x+1)~-~(x+h+1)}{(x+h+1)(x+1)}}{h} \\\\\\ \cfrac{\frac{x+1-x-h-1}{(x+h+1)(x+1)}}{h}\implies \cfrac{\frac{\underline{x+1}\underline{-x}-h\underline{-1}}{(x+h+1)(x+1)}}{h}\implies \cfrac{\frac{-h}{(x+h+1)(x+1)}}{h} [/tex]

[tex]\bf \cfrac{\frac{-h}{x^2+xh+2x+h+1}}{h}\implies \cfrac{-h}{x^2+xh+2x+h+1}\cdot \cfrac{1}{h} \\\\\\ \lim\limits_{h\to 0}~\cfrac{-1}{x^2+xh+2x+h+1}\implies \cfrac{-1}{x^2+x(0)+2x+0+1} \\\\\\ \lim\limits_{h\to 0}~\cfrac{-1}{x^2+2x+1}[/tex]