Answer:
We can use a numerical method such as the Newton-Raphson method or the bisection method to approximate the positive zero of the function f(x) = 2x - 16x³ - 3x² - 8x - 2.
Let's use the Newton-Raphson method to approximate the positive zero of f(x). We start by choosing an initial guess x_0, and then compute successive approximations using the formula:
x_(n+1) = x_n - f(x_n) / f'(x_n)
where f'(x) is the derivative of f(x). We continue this process until we get an approximation that is accurate enough for our needs.
Let's choose an initial guess of x_0 = 1.5. Then we have:
f(x) = 2x - 16x³ - 3x² - 8x - 2
f'(x) = 2 - 48x² - 6x - 8
Using these expressions, we can compute successive approximations as follows:
x_1 = x_0 - f(x_0) / f'(x_0) = 1.5 - (-23.375) / (-73) ≈ 1.320
x_2 = x_1 - f(x_1) / f'(x_1) = 1.320 - (-12.608) / (-50.673) ≈ 1.141
x_3 = x_2 - f(x_2) / f'(x_2) = 1.141 - (-5.364) / (-35.883) ≈ 1.067
x_4 = x_3 - f(x_3) / f'(x_3) = 1.067 - (-1.949) / (-31.120) ≈ 1.042
x_5 = x_4 - f(x_4) / f'(x_4) = 1.042 - (-0.361) / (-30.251) ≈ 1.029
x_6 = x_5 - f(x_5) / f'(x_5) = 1.029 - (-0.012) / (-30.055) ≈ 1.028
So the positive zero of f(x) is approximately 1.028, rounded to two decimal places.