1) We can use Newton's method. I assume the equation is f(x)=x3+x2+x+17=0 (not x3+x2+x+17=2). Differentiate this wrt x: f'(x)=3x2+2x+1. The formula is xn+1=xn-f(xn)/f'(xn). We need an arbitrary x0 to start the process. Let's start with x0=0, then x1=-17, x2=-11.4556..., x3=-7.7799..., x4=-5.3781..., x5=-3.8847..., x6=-3.0947..., x7=-2.8332..., x8=-2.8049..., x9=-2.8046..., x10=-2.804642..., x11=-2.804642727. The latter seems to be stable and x=-2.804642727 is the approximate solution.
2) In this method we first need to establish the function f(x)=2x-log(x)-6, but it's not clear what the log base is meant to be, so we'll use natural log and base 10 as two different bases. Whichever base we use log(1)=0, so f(1)=2-6=-4. We need a value of x such that f(x)>0. f(4)>0 whichever base is used. We now have two values of x where f(x) goes from negative to positive, which indicates that the solution (zero) for f(x) is between x=1 and x=4. We can work out the equation of the line joining f(1) to f(4).
a) Base 10.
f(1)=-4, f(4)=1.3979 approx. m (slope of line)=(f(4)-f(1))/(4-1)=1.7993 approx. The equation of the line is:
y+4=1.7993(x-1) by plugging in the slope and the point (1,-4). This line intersects the x-axis hopefully close to the root of f(x)=0. So when 4=1.7993(x-1), x=4/1.3979+1=3.2231. f(3.2231)=-0.06213 (which is indeed close to the actual zero but negative).
We can now create another line joining f(3.2231) to f(4), giving us the slope:
(f(4)-f(3.2231)/(4-3.2231)=1.8793. We continue like this to get the next secant line and finding the intersection with the x-axis. The computations are complicated, but, to summarise, we eventually get to x=3.256367 approx, which is the zero of the given equation.
b) Base e.
f(1)=-4, f(4)=0.6137 approx. m=(f(4)-f(1))/(4-1)=1.5379 approx. The equation of the line is:
y+4=1.53793(x-1) by plugging in the slope and the point (1,-4). When y=0 we get another x value which should be close to the actual zero of f(x). So when 4=1.5379(x-1), x=4/1.5379+1=3.6009. f(3.6009)=-0.07930 (close to the actual zero and negative).
We can now create another line joining f(3.6009) to f(4), giving us the slope:
(f(4)-f(3.6009)/(4-3.6009)=1.7366. The equation of the new line is:
y+0.07930=1.7366(x-3.6009) by plugging in the new slope and the point (3.6009,f(3.6009))=(3.6009,-0.07930). This line intersects the x-axis at 0.07930/1.7366+3.6009=3.6466. f(3.6466)=-0.000575 (close to the actual zero). Eventually we arrive at x=3.646945, the required zero.
