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# interval bisection watch

1. sin 3x = x^2 interval .6 < x < 1

rearrange to sin3x - x^2 = f(x)

So,

f(0.6)=sin(1.8) - .36 = -0.33
f(1)=sin(3) - 1 = -0.95

As there is no sign change present that implies there is no root between these two point but the answer says different?
2. (Original post by B.Carmine)
sin 3x = x^2 interval .6 < x < 1

rearrange to sin3x - x^2 = f(x)

So,

f(0.6)=sin(1.8) - .36 = -0.33
f(1)=sin(3) - 1 = -0.95

As there is no sign change present that implies there is no root between these two point but the answer says different?
3. (Original post by TeeEm)
Nope, and there's a note at the bottom of the page telling me just that, only got to there now ://////////////////////////////
5. Can anyone explain why it is radians? Or will it specify it in the question?
6. (Original post by B.Carmine)
Can anyone explain why it is radians? Or will it specify it in the question?
You cannot mix algebraic quantities with trigonometric quantities unless you are in radians.

As a rule of thumb at Advanced Maths use radians unless the question says to work in degrees.
7. (Original post by TeeEm)
You cannot mix algebraic quantities with trigonometric quantities unless you are in radians.

As a rule of thumb at Advanced Maths use radians unless the question says to work in degrees.

Ah that makes sense, alright cheers!
8. (Original post by B.Carmine)
Ah that makes sense, alright cheers!
my pleasure

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