Answer should be up to one decimal place only.

f(x) = x^2 - 2 on [0,3]
asked Aug 12 in Calculus Answers by skeptic (160 points)

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When x=0, f(0)=x^2-2=-2 and f(3)=9-2=7. So there is a change of sign between x=0 and x=3, implying that the root lies in between.

The average or midway position is x=1.5 and f(1.5)=2.25-2=0.25. Therefore the root lies between 1.5 and 0.

The midway position is 0.75 and f(0.75)=0.5625-2=-1.4375. So the root lies between 0.75 and 1.5, i.e., 1.125 as the midway point. And f(1.125)=-0.734375. The midway point including the root is (1.125+1.5)/2=1.3125.

f(1.3125)=-0.27734375, meaning that the root lies between 1.3125 and 1.5, with a midpoint of 1.40625, and f(1.40625)<0, so the root lies in the interval [1.40625,1.5], the midway point being 1.453125, and f(1,453125)>0 so the root lies in the interval [1.40625,1.43125], the midway point of which is 1.41875. We can see that to one decimal place the root is 1.4, because both extremes of the limit give us 1.4.

answered Aug 12 by Rod Top Rated User (479,220 points)
selected Aug 12 by skeptic
Thank you Sir Rod!

Hoping to help for my other post. Thanks again!
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