Question

Prove that b/w any two roots of f(x)=0, there is a root of g(x)=0 where (1) f(x)=e^xcos(x+1) , g(x)=e^x sinx+1

Answer 1

Answer:

0

Step-by-step explanation:

To elaborate on this a bit more:

excos(x)−1

Multiply everything through by e−x:

e−x(excos(x)−1)=cos(x)−e−x

Notice that h(x)=cos(x)−e−x and f(x)=excos(x)−1 have the same roots as h(x)=e−xf(x) and e−x is never 0. So now take the deriviative of h(x):

h′(x)=−sin(x)+e−x

Given any two consecutive roots of h (and, well f), say a and b, by Rolles Theorem there exists c such that h′(c)=0 so:

0=−sin(c)+e−c=sin(c)−e−c=ec(sin(c)−e−c)=ecsin(c)−1