For both the bisection and false position methods, which of the following applies? O a. If f(x₁)*f(x) > 0, then x₁ = x,
Posted: Thu May 05, 2022 5:37 pm
For both the bisection and false position methods, which of the following applies? O a. If f(x₁)*f(x) > 0, then x₁ = x, and x₁ stays the same. ○b. If f(x₁)*f(x) > 0, then xµ = (x₁ + x)/2 and x₁ = (x₁ + x₁)/2. Oc. If f(x₁)*f(x₁) < 0, then x₁ = x₁ and x₁ stays the same. Ⓒd. If f(x₁)*f(x) > 0, then xµ = (x₁ + x₁)/2 and x₁ = (x₁ + Xu)/2. ○e. If f(x₁)*f(x₁) > 0, then x = x₁ and x₁ stays the same. ○f._If f(x₁)*f(x) > 0, then xã = x, and x₁ stays the same. Og. If f(x₁)*f(x) < 0, then x₁ = x, and x₁ stays the same. ○h. If f(x₁)*f(x) < 0, then x = (x₁ + x₁)/2 and x₁ = (x₁+x₁)/2. x
When trying to find the root of y = x² - 6x + 9 = 0, which of the following methods will work to a precision of 1x10-6? a. the false position method but not the bisection method. Ob. only the secant method. O c. only the Newton-Raphson, secant and modified secant methods. O d. none of the methods as the function never touches the x axis. Oe. only the Newton-Raphson method. O f. all of the methods: bisection, false position, Newton-Raphson, secant and modified secant. g. only the bisection, false position and Newton-Raphson methods. Oh. only the bisection and false position methods. X
When trying to find the root of y = x² - 6x + 9 = 0, which of the following methods will work to a precision of 1x10-6? a. the false position method but not the bisection method. Ob. only the secant method. O c. only the Newton-Raphson, secant and modified secant methods. O d. none of the methods as the function never touches the x axis. Oe. only the Newton-Raphson method. O f. all of the methods: bisection, false position, Newton-Raphson, secant and modified secant. g. only the bisection, false position and Newton-Raphson methods. Oh. only the bisection and false position methods. X