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Bisection Method | Example 2 | Numerical Computation
Рет қаралды 21,193
3 жыл бұрын
@maskertravel8189 2 жыл бұрын
can you do for the question i. im stil stuck how to solve when the interval was 0
@aestheticart4955 Жыл бұрын
Why did you take the values to 4 decimal places?
@bharatinaikodi01 2 жыл бұрын
Thankyou For this Video, It was really informative and easy to understand and it really helped me to complete my assignment 😅
@theinfinitemath11 2 жыл бұрын
Thank you for the cheer… And you are most welcome…
@ppsasi3694 2 жыл бұрын
thank you so much for this wonderfull video
@theinfinitemath11 2 жыл бұрын
You are most welcome…
@maxima4116 2 ай бұрын
what i dont pick the (1, 3.2) interval, would it still yield the same result if i were to pick (0,1)?
@nthabisengmokoena5473 2 жыл бұрын
Why is it F(a) changing when f(x) is a positive? Shouldn't A change when f(x) is negative and B change when f(x) is positive?
@theinfinitemath11 2 жыл бұрын
Thank you for the comment! The root is where the graph is crossing the x-axis. So, it will always lie in an interval where the graph is going from positive to negative or negative to positive, y coordinates. When f(x) is positive it replaces the positive value on the y-axis, so that we still have both positive and negative values in the interval. Therefore: The positive value between f(a) and f(b) changes when f(x) is positive. And the negative value between f(a) and f(b) changes when f(x) is negative. Hope this helps…
@adoptedshrek6655 8 ай бұрын
Thank u mam'
@theinfinitemath11 8 ай бұрын
Most Welcome…
@wagayeweno54 2 жыл бұрын
It is not seen perfectly
@theinfinitemath11 2 жыл бұрын
Increase the picture quality of the video.
The Organic Chemistry Tutor
Рет қаралды 785 М.