(a) If there is a root of the equation ex+x=0
(b) The root of the equation to six decimal place.
(a) Let fx=ex+x. Intermediate Value Theorem states that if a continuous function f(x) on the interval [a,b] has values of opposite signs inside an interval, there must be some value x=c on the interval (a,b), for which fc=0.
f0=1>0 and f-1=e-1-1<0(because e-1<1). Since f is a continuous function on [-1,0], by Intermediate Value Theorem there exists c ϵ(-1,0) such that fc=0.
(b) From (a) we get
Let x0 be our initial estimate of the root, and let xn be the n-th estimation.f'x=ex+1, by Newton’s method we have xn+1=xn-fxnf'xn, xn+1=xn-exn+xnexn+1=xn-1exnexn+1.
Let x0=0, then x1=x0-1ex0ex0+1=0-1e0e0+1=-12.
Since x4=x5 to 8 decimal places, we can therefore hope that -0.56174329041 is close enough.
(a) Using the Intermediate Value Theorem, we get fc=0
(b) Using by Newton’s method we get the root of the equation as -0.56174329041