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Halving the interval to a required decimal place (1 Viewer)

mrpotatoed

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What exactly does it mean to one decimal place? Does it have to be the y-value of the function when an x value is tested? IE: F(2.125)=-0.002 which is to three DP, which is even better than one dp, does that mean that answer is x=2.125 as an approximation to 1dp? :blink2:
 

InteGrand

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What exactly does it mean to one decimal place? Does it have to be the y-value of the function when an x value is tested? IE: F(2.125)=-0.002 which is to three DP, which is even better than one dp, does that mean that answer is x=2.125 as an approximation to 1dp? :blink2:
By 1 decimal place, it means teh x-value approximation needs to be correct to 1 decimal place. In other words, keep getting better approximations and see when your root is trapped between two endpoints which, when rounded to 1 d.p., are the same. For example, if you can deduce the root is between x = 1.15 and x = 1.23, then you know to one decimal place, the root is going to be 1.2. (I just used random values here, wasn't referring to the specific question in the jpeg file).
 

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