How do you use the Intermediate Value Theorem to show that the polynomial function # [cos (t)] t^3 + 6sin^5(t) -3=0# has a root on the interval (0,2pi)?

Question

1788 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-17T16:19:57

Refer to explanation

Set #f(x)= (cos (t)) t^3 + 6sin^5(t) -3#.Then we have that

#f(0)=(cos0)*0+6*sin0-3=-3<0# nad

#f(2pi)=(cos2pi)*(2pi)+6sin^5(2pi)-3=2pi-3>0#

Because #f(0)*f(2pi)<0# the IVT tell us that f(x) has a root in the
interval #(0,2pi)#

Answer 2 · 2015-09-17T16:22:43

Lets calculate #y# for two values #0# and #pi/2#.

#y(0)=cos(0)*0^3+6*sin^5 0 - 3=1*0+6*0-3=-3#
#y(pi/2)=cos(pi/2)*(pi/2)^3+6*sin^5(pi/2)-3#
#y(pi/2)=0*(pi/2)^3+6*1^5-3=0+6-3=3#

So, using IVT, there is #c in (0,pi/2)sub(0,2pi)# such as #y(c)=0#