Question

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)

Answer 1

`\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{(1)/(x+h+1)-(1)/(x+1)}{h}\implies \cfrac{((x+1)~-~(x+h+1))/((x+h+1)(x+1))}{h} \\\\\\ \cfrac{(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{(-h)/((x+h+1)(x+1))}{h}`


`\bf \cfrac{(-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}`