Question

Let f(x) = x^2 + 1
lim h -> 0 [f(x+h) - f(x)]/h

Answer 1

\[\large \lim_{h \rightarrow 0} \frac{{(x+h)}-f(x)}{h}\] Given function \[\large f(x)=x^2+1\] plug your f(x) into your "x" value into the definition of a derivative.\[\large \lim_{h \rightarrow 0}\frac{[(x^2+1)+h]-(x^2+1)}{h}\] are you able to evaluate this now?

Answer 2

Made a little booboo Jhanny :P

Answer 3

Where at?

Answer 4

Oh i forgot the f in the definition?

Answer 5

(x+h)^2+1

Answer 6

ahh...yeah.

Answer 7

2x

Answer 8

\[\large \lim_{h \rightarrow 0} \frac{f{(x+h)}-f(x)}{h}\] Given function \[\large f(x)=x^2+1\] plug your f(x) into your "x" value into the definition of a derivative.\[\large \lim_{h \rightarrow 0}\frac{[(x+h)^2+1]-[x^2-1]}{h}\] are you able to evaluate this now?

Answer 9

omg it'ssupposed to be x^2+1 inside. -_- i hate latex :|

Answer 10

latex LOL

Answer 11

\[\large \lim_{h \rightarrow 0}\frac{[(x+h)^2+1]-[x^2+1]}{h}\]\[\large \lim_{h \rightarrow 0}\frac{x^2+2xh+h^2+1-x^2-1}{h}\]\[\large \lim_{h \rightarrow 0}\frac{2xh+h^2}{h}\]factor out an h. \[\large \lim_{h \rightarrow 0}\frac{\cancel h(2x+h)}{\cancel h}\] when h goes to 0, you get 2x as your answer.

Answer 12

thanks

Answer 13

np, understand how it works?

Answer 14

nope :)

Answer 15

yes