Halving the interval help (1 Viewer)

[Catberry]

How would I approach this question?

a)show that f(x)=2x^3+2x-1 hasa. Root between x=0 and x=1
b)by finding f'x, explain why this is the only root
c)use one application of halving the interval to find a smaller interval containing the root.

 

[asianese]

a) What is the condition for a root between two points? What needs to occur at x=0 and x=1?

b) Find f'(x)...what is it? The resulting equation is a qu___ and so if we find its d___....?

c) How do we halve the interval?

 

[bokat]

a) Sub zero for x in f(x) to get f(0)=-1
Sub one for x in f(x) to get f(1)=3
As f(x) changes from negative to positive between 0 and 1
so there is a root for f(x) between those values of x.

b)Differentiate f(x) to get f'(x)=6x^2+2. This is always positive
for all values of x which means f(x) is always increasing so it
can cross the x axis once only and will have only one zero.

c) Find the nimber half way between the given values of x which is (0+1)/2=1/2
Find f(1/2) which is 1/4.
As this is positive we can replace it with one which also made f(x) positive
so the new interval is between 0 and 1/2.
Hope this is useful.
Reza Bokat,
Cambridge Coaching
www.cambridgecoaching.com.au