Question

show that the root of quation x^3-6x+3=0 lies between 2 and 3x

Answer 1

Answer:

The three roots are: -0.670622, 0.756389, 5.91423.

Step-by-step explanation:

Let f(x) = x^3 - 3x^2 + 3.

f(-1) = -4; f(0) = 3; f(+1) = -2.

The function f(x) crosses the X-axis berween x = -1 and x = 0 with a +ve slope and again with a -ve slope between x = 0 and x = 1. By Bolzano’s theorem the cubic has two roots in the interval (-1, 1). f’(x) = 3(x)(x -4) = 0; x =0 and x =4 are points of optima. f”(x) = 6x-12. f”(0) = -12, at x=0, f(0) is a maximum. f”(4) = 12, at x = 4, f(4) = 64–96 +3 = -29 is a minimum. At x =6, f(6) = 216–216 +3 = 3, +ve. By Bolzano’s theorem, there is a third root in (4,6).

The three roots are: -0.670622, 0.756389, 5.91423. (Obtained using Wolfram-Alfa)