## 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.

1. anonymous

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

2. anonymous

3. anonymous

why is it 1,2 ?

4. anonymous

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

5. anonymous

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

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

7. anonymous

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

8. anonymous

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

9. anonymous

which explains the $f(t)-2$ in the numerator