anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
you are suposed to recognize this as the derivative of \[t^2+t\] at the point (1,2)
anonymous
  • anonymous
option 2 in your case
anonymous
  • anonymous
why is it 1,2 ?

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anonymous
  • anonymous
oh because the derivative at a point \[a\] is \[lim_{x->a} \frac{f(x)-f(a)}{x-a}\]
anonymous
  • anonymous
in this case i guess i should have used t instead of x, but it makes no difference. \[a=1\] \[f(1)=2\]
anonymous
  • anonymous
but howd u know whether to use a=1 or a= -1
anonymous
  • anonymous
so you have \[lim{t->1}\frac{f(t)-f(1)}{t-1}=lim_{t->1}\frac{f(t)-2}{t-1}\]\]
anonymous
  • anonymous
look at your denominator. it is\[t-1\] so this is the derivative at 1. also \[f(1)=2\]
anonymous
  • anonymous
which explains the \[f(t)-2\] in the numerator

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