Question

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

Answer 1

Here's one way to do it.

Explanation:

Let `f(x) = x^3-2x^2+3x`.
(Needed because the intermediate value theorem is a theorem about functions .)

Observe that the equation `x^3 - 2x^2 + 3x = 5` has a root (a solution) exactly when `f(x)=5`

So the question now is to show that for at least one number `c`, in `[1,2]`, we get `f(c)=5`.

`f` is continuous on `[1,2]` (Because it is a polynomial and they are continuous everywhere.)

`f(1) = 2` and `f(2) = 6`

`5` is between `f(1)` and `f(2)`, so

by the intermediate value theorem, there is at least one number `c`, in `[1,2]`, for which `f(c)=5`.

That is, the original equation has a solution.

Answer 2

Assuming you meant `f(x)=x^3-2x^2+3x-5`
Evaluate `f(1)` and `f(2)` and note that the function is continuous with values on both sides of `0`

Explanation:

Note:
`x^3-2x^2+3x=5` is a polynomial equation; it is not a polynomial function. Reference to an equation having a zero in an interval is meaningless.

`f(x)=x^3-2x^2+3x` does not have a zero in the interval `[1,2]`

That only leaves
`f(x)=x^3-2x+3x-5` as the intended function

`f(1)=-3`

`f(2)= +1`

Since `f(x)` is continuous in the interval `[1,2]` it must have have values for everything in the range `[-3,+1]` including zero.