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For part(d) shouldn't 1.89 also be a value?

Because u can do 6.92-2pi and u get a value of 0.64
Then u could add 1.25 right?


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Reply 1
Original post by Lilly1234567890
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For part(d) shouldn't 1.89 also be a value?

Because u can do 6.92-2pi and u get a value of 0.64
Then u could add 1.25 right?


Posted from TSR Mobile


What's the expression of the derivative that you get?

Edit: Moved to maths.
(edited 8 years ago)
It is plural in the question - stationary points - so there will be more than one x value :smile:
Original post by Zacken
What's the expression of the derivative that you get?

Edit: Moved to maths.


So I got
F'(X): 2+root10cos(X-1.25)


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Original post by enaayrah
It is plural in the question - stationary points - so there will be more than one x value :smile:


Yeah I got 3.50 and 5.28 which are correct but I also got a value of 1.89 however. This isn't the correct answer and I don't see why


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Original post by Lilly1234567890
So I got
F'(X): 2+root10cos(X-1.25)


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Original post by Lilly1234567890
Yeah I got 3.50 and 5.28 which are correct but I also got a value of 1.89 however. This isn't the correct answer and I don't see why


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I'd have thought you'd only get two solutions for cosx=-2/√10

But I'll let @Zacken explain to you nice and eloquently :yep:
Reply 6
Original post by Lilly1234567890
Yeah I got 3.50 and 5.28 which are correct but I also got a value of 1.89 however. This isn't the correct answer and I don't see why


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Bear with me, I'm in bed and the lights are off, so no pen and paper.

We have: cos(x1.25)=210\cos(x-1.25) = \frac{-2}{\sqrt{10}}, now our key-angle is arccos210\arccos{\frac{2}{\sqrt{10}}}, so the solutions are going to be given by (because the RHS is negative) x1.25=π+arccos210,πarccos210x-1.25 = \pi + \arccos \frac{2}{\sqrt{10}}, \pi - \arccos\frac{2}{\sqrt{10}}, then adding 1.25 to each. What do you propose that gets us 1.89?
Original post by Zacken
Bear with me, I'm in bed and the lights are off, so no pen and paper.

We have: cos(x1.25)=210\cos(x-1.25) = \frac{-2}{\sqrt{10}}, now our key-angle is arccos210\arccos{\frac{2}{\sqrt{10}}}, so the solutions are going to be given by (because the RHS is negative) x1.25=π+arccos210,πarccos210x-1.25 = \pi + \arccos \frac{2}{\sqrt{10}}, \pi - \arccos\frac{2}{\sqrt{10}}, then adding 1.25 to each. What do you propose that gets us 1.89?


oh yeah I did it again and i have no idea where i got the 1.89 from. oopsie
thanks though !
Reply 8
Original post by Lilly1234567890
oh yeah I did it again and i have no idea where i got the 1.89 from. oopsie
thanks though !


Sure thing.

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