Numerical analysis(Bisection Method)

[shayaan_musta]

My dear all!
Hello! How are you all?

Let us assume that we have given an equation of a curve like y=3x²+3x-1
By numerical analysis(bisection method) I want to find its root.
Everything is fine in the case if interval is given (a,b).
But my problem is that if interval is not given then what should I do? Means what should I taken as "a" or what should I taken as "b" to get first mid-point?
As I know that "a" & "b" have opposite signs.

Thanks in advance.

 

[hgfalling]

a and b don't have opposite signs (necessarily), the function evaluated at those points has opposite signs.

Now you can choose any points you want, as long as they satisfy that condition. In this case, you have a parabola. So one way to find points that will be suitable is to take the derivative of the function and set it to zero. This will find the min value of the function (or the max). Then you can just run along in one direction until you reach an x-value that flips the sign.

So here we have f'(x) = 6x + 3, so the min value of the parabola is at x=-1/2. f(-1/2) = -7/4 We can see that if x = 1, f(x) = 5. So those are two points.

But basically, you are just going to search around, either by plugging in, or some simple calculus, looking for a couple of points where the function changes sign.

 

[awkward]

It's usually a good idea to sketch the graph of the function.

 

[shayaan_musta]

a and b don't have opposite signs (necessarily), the function evaluated at those points has opposite signs.

Now you can choose any points you want, as long as they satisfy that condition. In this case, you have a parabola. So one way to find points that will be suitable is to take the derivative of the function and set it to zero. This will find the min value of the function (or the max). Then you can just run along in one direction until you reach an x-value that flips the sign.

So here we have f'(x) = 6x + 3, so the min value of the parabola is at x=-1/2. f(-1/2) = -7/4 We can see that if x = 1, f(x) = 5. So those are two points.

But basically, you are just going to search around, either by plugging in, or some simple calculus, looking for a couple of points where the function changes sign.

You are saying as one way to find points is to take derivative, so if you see without derivative if I put x=-1/2 then I get f(x)=-7/4 too.
So why take derivative if without & with derivative there is a same answer.

In this case, you have a parabola. So one way to find points that will be suitable is to take the derivative of the function and set it to zero.

If it is not a parabola then can't it give points without derivative?
I mean only a parabolic equation can give points at derivative? Or it is necessary to take derivative of a parabolic equation to find points?

 

[shayaan_musta]

It's usually a good idea to sketch the graph of the function.

Don't you think to plot a equation is a difficult task?
Even it is a function?

 

[hgfalling]

You are saying as one way to find points is to take derivative, so if you see without derivative if I put x=-1/2 then I get f(x)=-7/4 too.
So why take derivative if without & with derivative there is a same answer.


If it is not a parabola then can't it give points without derivative?
I mean only a parabolic equation can give points at derivative? Or it is necessary to take derivative of a parabolic equation to find points?

Taking the derivative just guarantees that you can easily find a minimum or maximum of the function. It helps most when it's not that obvious where a function is below or above zero.

It doesn't matter that it's a parabola, and you can do it equally by guessing, plotting the function, or other techniques.

 

[Mark44]

Don't you think to plot a equation is a difficult task?
Even it is a function?

Much of a standard course on precalculus functions is spent sketching the graphs of a wide variety of functions. For the function that you asked about, it is not at all difficult to sketch a graph. As already mentioned the graph of y = 3x² +3x - 1 is a parabola that opens upward.

 

[shayaan_musta]

Taking the derivative just guarantees that you can easily find a minimum or maximum of the function. It helps most when it's not that obvious where a function is below or above zero.

It doesn't matter that it's a parabola, and you can do it equally by guessing, plotting the function, or other techniques.

I got your point. Thanks.

 

[shayaan_musta]

Much of a standard course on precalculus functions is spent sketching the graphs of a wide variety of functions. For the function that you asked about, it is not at all difficult to sketch a graph. As already mentioned the graph of y = 3x² +3x - 1 is a parabola that opens upward.

May be you are a professional or may you are a genius therefore its easy to say for you.
Can you refer me any book by which I could be able to learn graph sketching?