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FP1

activity10.2_2a.jpg

2. Use the method of Interval Bisection twice to get approximate solutions to these equations, starting with the given intervals:

a) sin 3x = x2 - 1, -1 < x < 0

answer
2.

a) f(-0.25) = 0.256



....??????
Reply 1
f(x) = -x^2 + sin3x + 1
Original post by ckfeister
f(x) = -x^2 + sin3x + 1


f(0.5)0.2474≉0.256f(-0.5) \approx -0.2474 \not \approx 0.256

f(0.25)0.256≉0.38f(-0.25) \approx 0.256 \not \approx 0.38
Reply 3
Original post by RDKGames
f(0.5)0.2474≉0.256f(-0.5) \approx -0.2474 \not \approx 0.256

f(0.25)0.256≉0.38f(-0.25) \approx 0.256 \not \approx 0.38


How...
Original post by ckfeister
How...


Plug it into your calculator correctly...
Reply 5
Original post by RDKGames
Plug it into your calculator correctly...


I got the plus and minus at -x^2 confused, I did -0.5^2 not -(-0.5^2)
Original post by ckfeister
I got the plus and minus at -x^2 confused, I did -0.5^2 not -(-0.5^2)


Should be (0.52)-(-0.5^2), something to look out for more carefully in the future :smile:
(edited 7 years ago)
Original post by ckfeister
I got the plus and minus at -x^2 confused, I did -0.5^2 not -(-0.5^2)


Are you using degrees or radians? In general you use radians with this sort of question
Reply 8
Original post by alfredholmes
Are you using degrees or radians? In general you use radians with this sort of question


Radians, I've done it now.

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