Question

How do you find the exact relative maximum and minimum of the polynomial function of `g(x) =x^3 + 6x^2 – 36`?

Answer 1

See below.

Explanation:

Given
`g(x)=x^3+6x^2-36`

To find the maximum and minimum of the function find first derivative and set it equal to zero.
`g'(x)=3x^2+12x=0`

After factoring out `3x` we obtain
`3x(x+4)=0`

Since `3!=0`, therefore
`x=0, and x=-4` are the two points of inflection.

For `x=0,` we obtain `y=-36`, and for `x=-4`, we get `y=-4`

Hence `(0,-36) and (-4,-4)` are the points of interest.

Now to ascertain the second derivative of the given function
`g''(x)=6x+12`

At the first point `(0,-36)`
`g''(0)=6xx0+12=12`, a positive quantity. It is a local minimum for the value of `x`.

Similarly at the point `(-4,-4)`,
`g''(-4)=6xx(-4)+12=-12` a negative quantity. It is a local maximum for the value of `x`.

Verify by drawing a graph with the graphing tool
graph{y=x^3+6x^2-36 [-80, 80, -40, 40]}