Question

How do you find points of inflection and determine the intervals of concavity given `y=4x^3+21x^2+36x-20`?

Answer 1

There is an inflection point at `x=-1.75` and the function is concave down (`nn`) on the interval `(-oo,-1.75)`, and it is concave up (`uu`) on the interval `(-1.75,oo)`.

Explanation:

Concavity and inflection points of a function can be determined by looking at the second derivative. If the second derivative is 0, it is an inflection point (IE where the graph changes concavity). If the second derivative is positive, you know the graph is concave up (`uu`), and if the second derivative is negative, you know the graph is concave down (`nn`).

So let's start by finding the second derivative for our function,
`f(x)=4x^3+21x^2+36x-20`

`f'(x)=12x^2+42x+36`

`f''(x)=24x+42`

Now we can solve the following equation to find the inflection point(s):
`24x+42=0`

`24x=-42`

`x=-42/24=-1.75`

Now, let's look at the intervals before and after our only inflection point.

If we pick a point smaller than `-1.75`, say `-2`, and plug it into our second derivative, we get `-6`, which is negative, so our function must be concave down (`nn`) on the interval `(-oo,-1.75)`.

If we then pick a point larger than `-1.75`, say `0`, we get `42`, which is positive, so the function must be concave up (`uu`) on the interval `(-1.75, oo)`.

In conclusion, by analyzing the function's second derivative we figured out that it has an inflection point at `x=-1.75`, that it is concave down (`nn`) on the interval `(-oo,-1.75)`, and that it is concave up (`uu`) on the interval `(-1.75,oo)`.