Question

find the cube noot of 25 by bisection method uoto four places of decimal​

Answer 1

Step-by-step explanation:

Let f the function defined by: ∀x∈R,f(x)=x3−25. We know that f(3

Let f the function defined by: ∀x∈R,f(x)=x3−25. We know that f(3√

Let f the function defined by: ∀x∈R,f(x)=x3−25. We know that f(3√25

Let f the function defined by: ∀x∈R,f(x)=x3−25. We know that f(3√25)=0 and the function has only one root since it's strictly increasing and continuous.

Let f the function defined by: ∀x∈R,f(x)=x3−25. We know that f(3√25)=0 and the function has only one root since it's strictly increasing and continuous.f(0) for example is negative and f(3) is positive so you can consider the interval [0,3]. You can take any other numbers.

Let f the function defined by: ∀x∈R,f(x)=x3−25. We know that f(3√25)=0 and the function has only one root since it's strictly increasing and continuous.f(0) for example is negative and f(3) is positive so you can consider the interval [0,3]. You can take any other numbers.The number of iterations is as you wish. As much it's bigger, the value you get is closer to 3

Let f the function defined by: ∀x∈R,f(x)=x3−25. We know that f(3√25)=0 and the function has only one root since it's strictly increasing and continuous.f(0) for example is negative and f(3) is positive so you can consider the interval [0,3]. You can take any other numbers.The number of iterations is as you wish. As much it's bigger, the value you get is closer to 3√

Let f the function defined by: ∀x∈R,f(x)=x3−25. We know that f(3√25)=0 and the function has only one root since it's strictly increasing and continuous.f(0) for example is negative and f(3) is positive so you can consider the interval [0,3]. You can take any other numbers.The number of iterations is as you wish. As much it's bigger, the value you get is closer to 3√25

Let f the function defined by: ∀x∈R,f(x)=x3−25. We know that f(3√25)=0 and the function has only one root since it's strictly increasing and continuous.f(0) for example is negative and f(3) is positive so you can consider the interval [0,3]. You can take any other numbers.The number of iterations is as you wish. As much it's bigger, the value you get is closer to 3√25.

Answer 2

Answer:

find cube root of the 25 using bisection method correct to three decimal places