Interval Halving - Solve f(x)=x^3+2x^2+pi(x)-(square root of 2)

[ranger1716]

could someone give me a hand with this please?

i need to use the interval halving method to show that the function f has a root in the interval [a,b]. I need to approximate that root and determine a bound on the error of my estimate.

f(x)=x^3+2x^2+pi(x)-(square root of 2)

I have determined that f(-1)= -3.55581 and that

f(1)= 4.72738

therefore

(-3.55581)x(4.72738)<0

the actual answer to the problem in the back of the book is: root is approx=0.25 and error at most 1/8

I'm getting confused because wouldn't the error be (-1+1)/2? This would equal zero.

Any help would be great.

 

[CarlB]

The interval halving method requires you to keep narrowing an interval to find a smaller and smaller interval that contains a desired root. In this case, the intervale halving method goes like this:

-1) -3.55581
+1) +4.72738

So there must be a root between -1 and +1. Try the midpoint of the interval, i.e the midpoint of [-1,+1] or zero. This gives:

-1) -3.55581
0) -1.414
+1) +4.72738

So now you know that the solution is in the interval [0,+1]. You begin the problem again with:

0) -1.414
+1) +4.72738

So there must be a root somewhere between 0 and 1. Try 0.5
etc.

Carl