Halving interval (wording confusion) (1 Viewer)

[youngminii]

Let f(x) = x^3 + 5x^2 + 17x - 10. f(x) only has one root.
i) Show that root lies between 0 and 2
ii) Use one application of halving interval to find smaller interval containing root
iii) Which end of smaller interval in part ii) is closer to root? Briefly justify answer

The whole thing is worth 4 marks in 1995 3u HSC
I wanna know what you'd do for part iii)

 

[MPerk0]

dosnt it just mean if it is smaller than or greater than the answer you get from the first application of halving the interval? I.e in this case is it bigger or smaller than 0.5.

 

[xFusion]

yeh thats right. but you would say its equal to one or the other. then say because that answer is closer to zero so its a closer approx.

 

[Michaelmoo]

Ok Itd be better if we actually do the qestion for you to understand:

i) f(0) = -10 < 0
f(2) = 52 > 0

Therefore the root lies between x=0 and x=2 since f(0) and f(2) have opposite signs.

ii) f([0+2]/2) = f(1) = 13 > 0

Therefore the root lies between x = 0 and x=1

ii) Lets perform another application of halving the interval:

f([0+1]/2) = f(0.5) = -0.125

Hence the we know the root lies within the interval of x=0.5 to x=1. From this, the root is closer to x=1 than to x= 0


That would be enough to get you the marks if you show that working. If you really wanted to be fancy you could say:

"Since the function value x=1 remains in the interval range after one application of halving the interval, and x = 0 is replaced by a closer value x=0.5, then the root is closer to x=1 than to x=0.

 

[jet]

Ok Itd be better if we actually do the qestion for you to understand:

i) f(0) = -10 < 0
f(2) = 52 > 0

Therefore the root lies between x=0 and x=2 since f(0) and f(2) have opposite signs.

ii) f([0+2]/2) = f(1) = 13 > 0

Therefore the root lies between x = 0 and x=1

ii) Lets perform another application of halving the interval:

f([0+1]/2) = f(0.5) = -0.125

Hence the we know the root lies within the interval of x=0.5 to x=1. From this, the root is closer to x=1 than to x= 0


That would be enough to get you the marks if you show that working. If you really wanted to be fancy you could say:

"Since the function value x=1 remains in the interval range after one application of halving the interval, and x = 0 is replaced by a closer value x=0.5, then the root is closer to x=1 than to x=0.

Just make sure in part i) that you assert that the function is continuous on that interval. Don't just assume it, otherwise, your statement is not completely true in the mathematical sense. Otherwise, yeah, you need to do that second application.

 

[Michaelmoo]

Just make sure in part i) that you assert that the function is continuous on that interval. Don't just assume it, otherwise, your statement is not completely true in the mathematical sense. Otherwise, yeah, you need to do that second application.

Yep your right, I forgot about that step. You have to say the function is continuous for that domain. They're pretty strict on that.