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anonymous

  • 5 years ago

The following limit represents the derivative of some function f at some point (a, f(a)). Select an appropriate f(t) and a. http://www.webassign.net/cgi-bin/symimage.cgi?expr=lim_%28t-%3E1%29%20%28t%5E2%20%2B%20t%20-%202%29%2F%28t%20-%201%29 f(t) = t^2, a = 1 f(t) = t^2 + t, a = 1 f(t) = t - 2, a = 1 f(t) = t - 2, a = -1 f(t) = t^2 + t, a = -1 f(t) = t^2, a = -1 None of the other options is correct.

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  1. anonymous
    • 5 years ago
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    you are suposed to recognize this as the derivative of \[t^2+t\] at the point (1,2)

  2. anonymous
    • 5 years ago
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    option 2 in your case

  3. anonymous
    • 5 years ago
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    why is it 1,2 ?

  4. anonymous
    • 5 years ago
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    oh because the derivative at a point \[a\] is \[lim_{x->a} \frac{f(x)-f(a)}{x-a}\]

  5. anonymous
    • 5 years ago
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    in this case i guess i should have used t instead of x, but it makes no difference. \[a=1\] \[f(1)=2\]

  6. anonymous
    • 5 years ago
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    but howd u know whether to use a=1 or a= -1

  7. anonymous
    • 5 years ago
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    so you have \[lim{t->1}\frac{f(t)-f(1)}{t-1}=lim_{t->1}\frac{f(t)-2}{t-1}\]\]

  8. anonymous
    • 5 years ago
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    look at your denominator. it is\[t-1\] so this is the derivative at 1. also \[f(1)=2\]

  9. anonymous
    • 5 years ago
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    which explains the \[f(t)-2\] in the numerator

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