Question

How do you use the intermediate value theorem to explain why `f(x)=x^2-x-cosx` has a zero in the interval [0,pi]?

Answer 1

See explanation...

Explanation:

The intermediate value theorem tells us that if `f(x)` is a function continuous on an interval `[a, b]` and `y in [f(a), f(b)]` (or `y in [f(b), f(a)]` if `f(b) < f(a)`), then there is some `x in [a, b]` such that `f(x) = y`.

In our example:

`f(x) = x^2-x-cos x`

`f(x)` is continuous on `[0, pi]` (in fact on the whole of `RR`)

`f(0) = 0-0-cos(0) = -1 < 0`

`f(pi) = pi^2-pi-cos(pi) = pi(pi-1)+1 > 3(3-1)+1 = 7 > 0`

So `0 in [f(0), f(pi)]` and by the intermediate value theorem, there is some `x in [0, pi]` such that `f(x) = 0`