Question

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]?

Answer 1

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]}