# Solving equations numerically, C3 questionWatch

#1
I'm very much struggling with this chapter, I can get as far as getting the graph on the calculator and getting the table for the iteration stuff, but I mainly don't know what to do!

This is one of the questions I need help with:

(a) Given that f(x)= (e^2x) - 6x, find f(0)
plugged it into calculator and got 1

Evaluate f(1), correct to 3dp
Again, calculatored it and got 1.389

Explain how f(x)= 0 could still have a root in the interval 0< x< 1 even though f(0)f(1) > 0

(b) Rewrite the equation f(x)= 0 in the form x= F(x), for some suitable function of F.

Taking x0=0.5 as an initial approximation, use an iterative method to determine one of the roots of this equation correct to 3dp. How could you demonstrate that this root has the required degree of accuracy?
I don't know!

(c) Deduce the value to 2dp of one of the roots of the equation (e^x) - 3x
Alright so I get the graph on my calculator which tells me the roots are 0.62 and 1.51. Is this the way I'm supposed to do it?

Thanks for reading this, any help will be appreciated!
0
9 years ago
#2
Have you actually read the material in the chapter about solving equations numerically?
0
#3
I have indeed! I just find it really difficult, which is why I'm asking for help :/ And maths texbooks are hard to read
0
9 years ago
#4
(Original post by satsumaree)
I'm very much struggling with this chapter, I can get as far as getting the graph on the calculator and getting the table for the iteration stuff, but I mainly don't know what to do!

This is one of the questions I need help with:

(a) Given that f(x)= (e^2x) - 6x, find f(0)
plugged it into calculator and got 1

Evaluate f(1), correct to 3dp
Again, calculatored it and got 1.389

Explain how f(x)= 0 could still have a root in the interval 0< x< 1 even though f(0)f(1) > 0

(b) Rewrite the equation f(x)= 0 in the form x= F(x), for some suitable function of F.

Taking x0=0.5 as an initial approximation, use an iterative method to determine one of the roots of this equation correct to 3dp. How could you demonstrate that this root has the required degree of accuracy?
I don't know!

(c) Deduce the value to 2dp of one of the roots of the equation (e^x) - 3x
Alright so I get the graph on my calculator which tells me the roots are 0.62 and 1.51. Is this the way I'm supposed to do it?

Thanks for reading this, any help will be appreciated!
[B](b) is just asking you to rearrange f(x) to get an x on it's own. Once found, press 0.5 then equals on your calculator and sub ans as x into your equation, then keep pressing equals and you could get a root (or maybe not, if it doesn't work try another rearrangement)
0
9 years ago
#5
(Original post by DFranklin)
Have you actually read the material in the chapter about solving equations numerically?
Will this Rearrangement method work?

i have rearranged the original f (x)= x^3 + x - 5 to be y= 3√(5-x)!

i got 1.51598 to 5 sig figures to be the root and used 1 as the starting point. But when i use autograph solve y= 3√(5-x) along with a y=x graph i get different roots and converges to 1.057?

Also when both graphs are drawn do i use the x value where the graph croxes the x-axis or where the graphs intersect?

I know for a fact that if i rearrange y= 3√(5-x) to y= 5 - x^3 it will definately fail. So i'm half way there
Thanks
0
9 years ago
#6
[B](b) is just asking you to rearrange f(x) to get an x on it's own. Once found, press 0.5 then equals on your calculator and sub ans as x into your equation, then keep pressing equals and you could get a root (or maybe not, if it doesn't work try another rearrangement)
Will this Rearrangement method work?

i have rearranged the original f (x)= x^3 + x - 5 to be y= 3√(5-x)!

i got 1.51598 to 5 sig figures to be the root and used 1 as the starting point. But when i use autograph solve y= 3√(5-x) along with a y=x graph i get different roots and converges to 1.057?

Also when both graphs are drawn do i use the x value where the graph croxes the x-axis or where the graphs intersect?

I know for a fact that if i rearrange y= 3√(5-x) to y= 5 - x^3 it will definately fail. So i'm half way there
Thanks
0
9 years ago
#7
(Original post by Owais B)
Will this Rearrangement method work?

i have rearranged the original f (x)= x^3 + x - 5 to be y= 3√(5-x)!

i got 1.51598 to 5 sig figures to be the root and used 1 as the starting point. But when i use autograph solve y= 3√(5-x) along with a y=x graph i get different roots and converges to 1.057?

Also when both graphs are drawn do i use the x value where the graph croxes the x-axis or where the graphs intersect?

I know for a fact that if i rearrange y= 3√(5-x) to y= 5 - x^3 it will definitely fail. So i'm half way there
Thanks
Is that suppose to be the cubed root of (5-x)? If so, I don't think that's right because you'd get x^(1/3) = (5-x)^(1/3)...

& you draw the rearranged equation, and see where it intersects the line y=x.

Sorry if that's wrong, I can't remember fully
0
9 years ago
#8
Is that suppose to be the cubed root of (5-x)? If so, I don't think that's right because you'd get x^(1/3) = (5-x)^(1/3)...

& you draw the rearranged equation, and see where it intersects the line y=x.

Sorry if that's wrong, I can't remember fully
Thanks for replying , i asked my teacher and he couldn't also see why there were 2 different roots lol..but i just changed the 5 in the equation to a 4 and worked! and your right about needing a single equation for both failure and success!

I have 1 final question lol..umm for the rearragement method, how would you know whether the numerical value of the gradient of y= g (x) is more or less than one. And the new root that i found was 1.3788? It says you have to explain and i'm not really sure. how would you prove that when it fails, the gradient of f(x) at the root you are trying to find is greater than 1.
0
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