use the intermediate value theorem to show that a solution exist for

Sin(x)=x-1 in the interval [0,pi/2]

Question

use the intermediate value theorem to show that a solution exist for

Sin(x)=x-1 in the interval [0,pi/2]

zarkon · Answer

doesn't look true

anonymous · Answer

Just had an exam I put that it isn't true
but I am not sure

myininaya · Answer

Let $$f(x)=\sin(x)-x+1$$
$$f(0)=\sin(0)-0+1=1>0$$
$$f(\frac{\pi}{2})=\sin(\frac{\pi}{2})-\frac{\pi}{2}+1=2-\frac{\pi}{2}>0$$


we could try a smaller interval
but it is true

zarkon · Answer

it's not true...the root is at around 1.93

anonymous · Answer

right and we were not supposed to use calculators.

myininaya · Answer

graphing f(x)=sin(x)-x+1 i have an x-intercept on the interval [0,pi/2]

anonymous · Answer

lol I put that [0,pi/2] on the side just in case

myininaya · Answer

x=1.0177624

anonymous · Answer

this question was worth 20 pts maybe I'll get 5 for trying

zarkon · Answer

are you sure about that x-intercept

myininaya · Answer

plug it in
it works

myininaya · Answer

i mean its an approximation
so you should get approximately the same thing on both sides

zarkon · Answer

http://www.wolframalpha.com/input/?i=plot [Sin%28x%29-x%2B1]+x%3D0+to+x%3Dpi%2F2

anonymous · Answer

okay this is what I did. I used 0 and plugged it in to the equation and got -1 and than I plugged pi/2 and got stuck

across · Answer

Do you know what the IVT (intermediate value theorem) is? If so, then notice that you're trying to solve$$\sin(x)-x+1=0.$$You're given the range$$[0,\frac{\pi}{2}].$$What are the values of this function at the boundaries?$$f(0)=\sin(0)-(0)+1=1,$$$$f(\frac{\pi}{2})=\sin(\frac{\pi}{2})-(\frac{\pi}{2})+1=2-\frac{\pi}{2}.$$However,$$[1,2-\frac{\pi}{2}]$$is a range that doesn't include 0. Therefore, how can it have a solution in that interval?

Perhaps I'm doing something wrong.

The solution is at x ≈ 1.93456, which is greater than pi/2 ≈ 1.57079.

myininaya · Answer

then my calc is not working

zarkon · Answer

The IVT is inconclusive for this problem.

myininaya · Answer

graph sin(x)-x+1 in your calculator zarkon

myininaya · Answer

the ti83 plus

zarkon · Answer

it does not tell us that we do not have a root (even though we don't)

zarkon · Answer

are you in radians

myininaya · Answer

darn it 
i will never tell you

zarkon · Answer

lol

myininaya · Answer

ok you win
you got 20/20

zarkon · Answer

I would look at the derivative of f

zarkon · Answer

f'(x)=cos(x)-1<0 on [0,pi/2]

myininaya · Answer

i told my calc to always be in radians why is not a faithful companion

zarkon · Answer

f(pi/2)>0 thus there can be no root

zarkon · Answer

bad calc...bad calc