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anonymous

  • 4 years ago

lim h → 0 (f(2+h)-f(2))/h F(x)=x^2+1

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  1. anonymous
    • 4 years ago
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    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

  2. anonymous
    • 4 years ago
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    Could you please show work on how to solve this using the equation, so I know for future refrences?

  3. anonymous
    • 4 years ago
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    \[f(2)=2^2+1=5\] \[f(2+h)=(2+h)^2+1=4+4h+h^2+1=5+4h+h^2\] \[f(2+h)-f(2)=5+4h+h^2-5=4h+h^2\] \[\frac{f(2+h)-f(2)}{h}=\frac{4h+h^2}{h}=\frac{h(4+h)}{h}=4+h\]

  4. anonymous
    • 4 years ago
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    now you can take the limit as h goes to zero and you get 4 for sure

  5. anonymous
    • 4 years ago
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    all steps are there, nothing left out

  6. anonymous
    • 4 years ago
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    WOW thank you!

  7. anonymous
    • 4 years ago
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    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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