Question

How do you use the Intermediate Value Theorem to show that the polynomial function `P(x) = x^3 - 2x^2 - 5` has a zero in the interval [-1, -2]?

Answer 1

See the explanation, please.

Explanation:

There is no interval, `[-1,-2}`.

There is an interval `[-2,-1[]`, but this function does not have a sero in that interval.

It does have a zero, but the zero if in the interval `[2,3]` (I assume we want an interval using integer endpoints).

`P` is a polynomial function, so it is continuous at all real numbers. In particular, `p` is continuous on the interval `[2,3]`.

`P(2) = 8-8-5 = -5`

`P(3) = 27-18-5 = 4`

`0` is between `P(2)` and `P(3)`.

So, by IVT, there is a `c` in `(2,3)` for which `P(c) = 0`

Note Irrelevant for this particular problem, but important for understanding the import of IVT:

`1` is also between `P(2)` and `P(3)` (and `P` is continuous on `[2,3]`),
so there is a `c` in `(2,3)` at which `P(c) =1`

And `27/5` is between `P(2)` and `P(3)` (and `P` is continuous on `[2,3]`),
so there is a `c` in `(2,3)` at which `P(c) =27/5`

And so on for every number between `-5` and `4`.

Note 2

Because this is not my first rodeo, before answering the question, I looked at the graph of `P(x)`:

graph{y = x^3-2x^2-5 [-13.07, 18.99, -9.87, 6.15]}

Clearly this cubic has only one real zero and it is NOT negative.
(Descartes' rule of signs also tells us that there is no positive zeros for this polynomial.)