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
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
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,
In particular, there is some value
graph{((x1)^2+(ysqrt(2)+5)^20.01)((x3)^2+(ysqrt(2)+1/3)^20.01)(y+4/xtan((pix)/8)) = 0 [8.94, 11.06, 6.24, 3.76]}