Question

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.

Answer 1

you are suposed to recognize this as the derivative of \[t^2+t\] at the point (1,2)

Answer 2

option 2 in your case

Answer 3

why is it 1,2 ?

Answer 4

oh because the derivative at a point \[a\] is
\[lim_{x->a} \frac{f(x)-f(a)}{x-a}\]

Answer 5

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

Answer 6

but howd u know whether to use a=1 or a= -1

Answer 7

so you have
\[lim{t->1}\frac{f(t)-f(1)}{t-1}=lim_{t->1}\frac{f(t)-2}{t-1}\]\]

Answer 8

look at your denominator. it is\[t-1\] so this is the derivative at 1. also \[f(1)=2\]

Answer 9

which explains the \[f(t)-2\] in the numerator