Given f(x)=sinx, use the tangent line at x=1 to estimate f(1.1).
A. 0
B. 0.540
B. 0.841
D. 0.896
E. 1

Question

Given f(x)=sinx, use the tangent line at x=1 to estimate f(1.1). 
A. 0
B. 0.540
B. 0.841
D. 0.896
E. 1

jdoe0001 · Answer

use the tangent line   <---   meaning the derivative of the function

darkigloo · Answer

f '(x) = cosx

jdoe0001 · Answer

yeap

darkigloo · Answer

so what do i do with that?

jdoe0001 · Answer

so I gather.... you could just use f(1.1) = sin(1.1)   and get it

but you're asked to get it the same value, by means of the derivative
thus f'(1) = cos(1)

jdoe0001 · Answer

hmmm wait.. is not the same value

anonymous · Answer

The linear approximation of $f$ at some point $x_0$ is: $$
L(x) -f(x_0) = f'(x_0)(x-x_0)
$$

anonymous · Answer

In this case, $x_0 = 1$ and $x = 1.1$.

anonymous · Answer

Do you get it, or do I have to continue?

darkigloo · Answer

so is this how it should look? 
L(1.1) - sin(1) = cos(1) (1.1-1)

anonymous · Answer

Yes, and they want you to solve for $L(1.1)\approx f(1.1)$.

anonymous · Answer

Though to be honest, this question is a bit poor because there is nothing that makes  $\sin(1)$ any easier to calculate than $\sin(1.1)$.  It would have been better if they had you use $\sin(\pi)$ to approximate $\sin(3)$ or something.

darkigloo · Answer

oh ok. i got 0.896 as my answer.

anonymous · Answer

Linear approximation: http://www.wolframalpha.com/input/?i=cos%281%29+*+%281.1-1%29+%2B+sin%281%29 Actual value: http://www.wolframalpha.com/input/?i=sin(1.1)

darkigloo · Answer

ahh i understand. thanks