SOLUTION: Here's my questin Prove a that a root exists for the given function p(x)=60x(1+x)^72-(1+x)^72+1=0 im assuming you could find the exact value but since they dont ask for th

Algebra ->  Equations -> SOLUTION: Here's my questin Prove a that a root exists for the given function p(x)=60x(1+x)^72-(1+x)^72+1=0 im assuming you could find the exact value but since they dont ask for th      Log On


   



Question 104461: Here's my questin
Prove a that a root exists for the given function
p(x)=60x(1+x)^72-(1+x)^72+1=0
im assuming you could find the exact value but since they dont ask for the
exact value we would just need to show a solution where p(x)> 0 and where p(x)<0 to prove there is infact a root between the two intervals for ie using the IVT
thanks to any one who can help.

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
p%28x%29=60x%281%2Bx%29%5E72-%281%2Bx%29%5E72%2B1=0
Let's move some terms around and discuss,
p%28x%29=%281%2Bx%29%5E72%2A%2860x-1%29%2B1
Since %281%2Bx%29%5E72 is always positive or zero, let call that A(x) where A(x) is always positive except when x=-1 and A(-1)=0.
p%28x%29=A%28x%29%2A%2860x-1%29%2B1
Let’s talk about this function for very large positive and negative x.
For even modest values of x (x<-2 and x>0), A(x) grows rapidly because of the power of 72.
For large positive x, A(x) gives a very large, positive number, (60x-1) gives a large, positive number, therefore p(x) is a very large, positive number since positive times a positive equals a positive.
For large negative x, A(x) is a very large, positive number, (60x-1) is a large negative number, therefore p(x) is a very large, negative number since positive times a negative equals a negative.
Therefore between large positive x and large negative x comes a point where p(x)=0, that is, it has a root.
Or using the vocabulary of the IVT,
Since
lim%28+x-%3Einfinity%2C+p%28x%29+%29+%0D%0A=+infinity
lim%28+x-%3E-infinity%2C+p%28x%29+%29+%0D%0A=+-infinity
There exists an x in the interval from -infinity to infinity where p%28x%29=0.