#### To determine

**To find**:

(a) If there is a root of the equation ex+x=0

(b) The root of the equation to six decimal place.

#### Answer

(a) fc=0

(b) -0.56174329041

#### Explanation

(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

fx=ex+x.

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.

x2=x1-1ex1ex1+1=-12-1e-12 e-12+1=-32(e-12)e-12+1=-1.5e0.5e-0.5+1=-1.5*0.606530651.60653065=-0.56631100319.

x3=x2-1ex2ex2+1=-1.56631100319*0.5676155131.567615513=-0.56714316503.

x4=x3-1ex3ex3+1=-1.56714316503*0.567143361511.56714336151=-0.56714329041.

x5=x4-1ex4ex4+1=-1.56714329041*0.567143290411.56714329041=-0.56714329041.

Since x4=x5 to 8 decimal places, we can therefore hope that -0.56174329041 is close enough.

**Conclusion:**

(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