Question

PLEASE HELP picture included

Answer 1

I think that the istantaneous rate of change is the first derivative of such functions

Answer 2

namely, we have to compare this two functions:
\[f'\left( x \right) = \sin \left( {2x} \right),\quad g'\left( x \right) = x\]

Answer 3

isn't f'(x)= to 2cos(x)sin(x)?

Answer 4

and g'(x)=1?

Answer 5

yes! and \(2 \sin x \cos x= \sin(2x)\)

Answer 6

i never knew that thanks for that

Answer 7

Help http://openstudy.com/study#/updates/56509396e4b045175944f394

Answer 8

no, please we have this:
\[g\left( x \right) = \frac{{{x^2}}}{2} \Rightarrow g'\left( x \right) = x\]

Answer 9

by using quotient rule

Answer 10

namely?

Answer 11

i dont understand what you mean

Answer 12

we have to search for the right \(x\) value, namely we have to search for the value \(x_0\), such that, the subsequent condition holds:
\[\sin \left( {2{x_0}} \right) > {x_0}\]

Answer 13

of course, we have to refer to the values listed in your exercise

Answer 14

i see

Answer 15

please, if you use a calculator, keep in mind that you have to set the "rad" mode, not "deg" mode

Answer 16

"rad" stands for radians

Answer 17

i put it on rad now

Answer 18

ok! Now try, for example the first value: \(x=-0.8\)

Answer 19

is \(x=-0.8\) the right value, please?

Answer 20

do i just plug it into both?

Answer 21

yes!

Answer 22

thats pretty simple actually

Answer 23

then whichever one lets f(x) have a greater i r o c is the correct one right?

Answer 24

you have to compare these quantites:
\(\sin(-1.6)\) and \(x_0=-0.8\)

Answer 25

instantaneous R.O.C.****

Answer 26

yes!

Answer 27

Thank you SO much, you have no idea how much you've helped me today

Answer 28

thanks! :)
I have solved your exercise. Please what is the right option?

Answer 29

c) 0.9 and that makes sense because sin cannot be less than -1 and greater than 1

Answer 30

correct!
since: \(\sin(2 \cdot 0.9)=0.973>0.9\)

Answer 31

thank you :)

Answer 32

:)