Use Regular False method to determine where the smallest positive root is.
Choose two points a and b such that f(a) and f(b) are of opposite signs.
With a=0 and b=1 find the (first) value of point of approximation:
Therefore, the interval of unit length that contains the smallest positive root is [0;1].
Starting with this interval, find an interval of length 0.05 or less that contains the root, by Bisection method
Consider our interval "[x_L;x_R]=[0;1]". Find the center of the interval and the value of the function at this point:
Check at what part of the interval - "[x_L;x_C]" or "[x_c;x_R]" - there is a root: if the values of the function at the ends of one of these intervals are opposite, there is a root:
"f(0)f(0.5)<0,\\\\f(0.5)f(1)>0."
Hence, we will only consider "[0;0.5]". Perform the next iteration by bisection method:
Hence, we will only consider [0.25;0.5]. Perform the next iteration:
Hence, we will only consider [0.375;0.5]. Perform the next iteration:
Hence, we will only consider [0.438;0.5]. Perform the next iteration:
"x_c=\\frac{0.438+0.5}{2}=0.469,\\\\\nf(x_c)=0.0034.\\\\\n\\text{Is the root on the left of } x_c\\text{ or on the right?}\\\\\nf(0.438)f(0.469)>0,\\\\f(0.469)f(0.5)<0."
But the length of both of these intervals is less than 0.05, therefore, the smallest positive root lies between 0.469 and 0.5.
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