Question

Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answer to two decimal places.) tan x ≈ x

Answer 1

So you want the magnitude of the difference of tan(x) and x to be less than or equal to 0.1 .

Answer 2

less than

Answer 3

like the closest one to 0.1 but no 0.1. like 0.09

Answer 4

\[|\tan(x)-x| \le 0.1 \\ -0.1 \le \tan(x)-x \le 0.1 \\ x-0.1 \le \tan(x) \le x+0.1 \]
so you are saying within doesn't include .1 ?

Answer 5

yes, it is not include it.

Answer 6

I did that same thing but like i dont know what to do then

Answer 7

\[|\tan(x)-x| < 0.1 \\ -0.1 < \tan(x)-x < 0.1 \\ x-0.1 <\tan(x) < x+0.1\]

Answer 8

because a=0 so i tried [-0.1 , 0.1] and it told me it doesn't include 0.1

Answer 9

yeah I don't know either at this second

Answer 10

Do you have chegg?

Answer 11

i don't know what chegg is
but I know we should at least look in the interval (-pi/2,pi/2)

Answer 12

since at x=-pi/2 and x=pi/2 for y=tan(x) we have vertical asymptotes...

Answer 13

\[\frac{-\pi}{2} < \tan(x)<\frac{\pi}{2}\]

Answer 14

yeah i tried that. its incorrect

Answer 15

Well I'm not done playing with this idea
I

Answer 16

i am here because i am desesperate haha i have tried everything i know

Answer 17

If you are allowed to use a graphing calculator, click on the link and then click on the graph to enlarge it: http://awesomescreenshot.com/09d3p1fv1b

Answer 18

I am not supposed to.. but i mean, everyone tells me its the only way! Thanks, i will try that and let you know

Answer 19

Rounded to two decimal places: (-0.63, 0.63)

Answer 20

It is correct! Thank you so much, you just literally saved my life haha. BTW, what is that website with that graphic calc. I want it

Answer 21

www.desmos.com

Answer 22

Glad that was accepted. You are welcome.