Question

How do you find the exact relative maximum and minimum of the polynomial function of `f(x) =2x^3-3x^2-12x`?

Answer 1

Maximum `=(-1,7)`
Minimum `=(2,-20)`

Explanation:

Since the graph is not limited by a definition, we can just look for the extremal points through the dervivate of the function.

`2x^3-3x^2-12x`

Dervivating the function with rule `(x^n)'=nx^(n-1)`, will give.

`(2x^3-3x^2-12x)'=6x^2-6x-12`

After finding the `f'(x)` we can look for zero points of that graph. Because this will be equal to the extremal points.

`6x^2-6x-12=0`

From here we can take several paths. Such as either use the quadratic equation or factorize to find the zero points. This will give us.

`x=2` and `x=-1` which can be seen as `(x-2)(x+1)`.

We can now put these numbers into the original function `f(x)` and find the `y-`values of the extremal points.

`f(2)=2*2^3-3*2^2-12*2=-20`
`f(-1)=2*(-1)^3-3*(-1)^2-12*(-1)=7`

These will be equal to coordinates, and we can with logic say which is which by this calculation. Since `f(2)=-20` we know that this has to be equal to the minimum point. And that `f(-1)=7` has to be equal to the maximum point.

So by that, we can say that.

Maximum `=(-1,7)`
Minimum `=(2,-20)`