anonymous
  • anonymous
lim h → 0 (f(2+h)-f(2))/h F(x)=x^2+1
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
4 and in a week you will do this problem in your head in two seconds by saying "the derivative of \[x^2+1\] is \[2x\] and \[2\times 2=4\] but you have not gotten there yet so you have some work to do
anonymous
  • anonymous
Could you please show work on how to solve this using the equation, so I know for future refrences?
anonymous
  • anonymous
\[f(2)=2^2+1=5\] \[f(2+h)=(2+h)^2+1=4+4h+h^2+1=5+4h+h^2\] \[f(2+h)-f(2)=5+4h+h^2-5=4h+h^2\] \[\frac{f(2+h)-f(2)}{h}=\frac{4h+h^2}{h}=\frac{h(4+h)}{h}=4+h\]

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anonymous
  • anonymous
now you can take the limit as h goes to zero and you get 4 for sure
anonymous
  • anonymous
all steps are there, nothing left out
anonymous
  • anonymous
WOW thank you!
anonymous
  • anonymous
yw notice that the 5's add to zero. it will always work this way if the limit exists. because you have an h in the denominator, the only way for this limit to exist is if the constant goes and you have nothing but h's

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