Question

How do you verify the intermediate value theorem over the interval [0,3], and find the c that is guaranteed by the theorem such that f(c)=4 where `f(x)=x^3-x^2+x-2`?

Answer 1

`f` is a polynomial so `f` is continuous on the interval `[0,3]`.

`f(0) = -2` and `f(3) = 19`

`4` is between `f(0)` and `f(3)` so IVT tells us that there is a `c` in `(0,3)` with `f(c) = 4`

Finding the `c` requires solving

`x^3-x^2+x-2 = 4`. Which is equivalent to

`x^3-x^2+x-6 = 0`

Possible rational zeros are `+-1`, `+-2`, `+-3`, and `+-6`.

Testing shows that `2` is a solution.

So `c = 2`

Answer 2

The value of `c in (0, 3)` and is `c=2`

Explanation:

The intermediate value theorem states that if `f(x)` is a continuous function on the interval `[a,b]`, then there is a number `p` between `f(a)` and `f(b)`, `(f(a)!=f(b))` such that there is a number `c in (a,b)` such that `p=f(c)`

Here, `f(x)=x^3-x^2+x-2`, which is a polynomial function continuous on the Interval `[0,3]`

`f(0)=0-0+0-2=-2`

`f(3)=3^3-3^2+3-2=27-9+3-2=19`

`f(c) in (f(0, f(3)))`

`f(c)=4` such that `c in (0,3)`

Therefore,

`f(c)=c^3-c^2+c-2=4`

`=>`, `c^3-c^2+c-6=0`

`f(2)=2^3-2^2+2-2=8-4+2-2=4`