Show that #P(x)=3x^5-4x^4+2x^3-4x^3+2x-1# has a root in the interval #(1,2)#?

1 Answer
Jan 5, 2018

Please see below.

Explanation:

The intermediate value theorem states that if a continuous function #f(x)#, in an interval #[a,b]# as its domain, takes values #f(a)# and #f(b)# at each end of the interval, then it also takes any value between #f(a)# and #f(b)# at some point within the interval.

Note that the given function #P(x)=3x^5-4x^4+2x^3-4x^3+2x-1# is continuous.

As #P(x)=3x^5-4x^4+2x^3-4x^3+2x-1#

at #x=1# we have #P(1)=3-4+2-4+2-1=-1#

and at #x=2# we have #P(1)=3xx2^5-4xx2^4+2xx2^3-4xx2^2+2xx2-1#

= #96-64+16-16+4-1=35#

As #P(x)=3x^5-4x^4+2x^3-4x^3+2x-1# is defined in the interval #[1,2]# and is also continuous,

and it takes a value #-1# at #x=1# and #35# at #x=2#, it takes a value #0# somewhere between #1# and #2#. As such it crosses #x#-axis between #1# and #2#.