How do you use the intermediate value theorem to explain why #f(x)=-4/x+tan((pix)/8)# has a zero in the interval [1,3]?

1 Answer
Nov 30, 2016

Answer:

See explanation...

Explanation:

Given:

#f(x) = -4/x+tan((pix)/8)#

Note that:

  • The function #-4/x# has a vertical asymptote at #x=0#, but is otherwise defined and continuous. So is defined and continuous on #[1, 3]#

  • The function #tan theta# is defined for all #theta in (-pi/2, pi/2)#, including all #theta in [pi/8, (3pi)/8]#, and is continous in that interval.

Hence #f(x)# is defined and continuous over the interval #x in [1, 3]#

We find:

#f(1) = -4/1+tan(pi/8) = -4+(sqrt(2)-1) = sqrt(2)-5 < 0#

#f(3) = -4/3+tan((3pi)/8) = -4/3+(sqrt(2)+1) = sqrt(2)-1/3 > 0#

In summary:

  • #f(x)# is defined and continuous on the interval #[1, 3]#

  • #f(1) < 0# and #f(3) > 0#

So by the intermediate value theorem, #f(x)# takes every value between #f(1) = sqrt(2)-5 < 0# and #f(3) = sqrt(2)-1/3 > 0# for some #x in [1, 3]#.

In particular, there is some value #a in [1, 3]# such that #f(a) = 0#

graph{((x-1)^2+(y-sqrt(2)+5)^2-0.01)((x-3)^2+(y-sqrt(2)+1/3)^2-0.01)(y+4/x-tan((pix)/8)) = 0 [-8.94, 11.06, -6.24, 3.76]}