Question

How do you find points of inflection and determine the intervals of concavity given `y=x^4+x^3-2x^2+1`?

Answer 1

Use the second derivative test.

Explanation:

You can determine the intervals of concavity and inflection points of a function using the second derivative test. The second derivative test tells us whether the slope of the function is increasing or decreasing; when the second derivative is positive, the function is concave up, and, similarly, when the second derivative is negative, the function is concave down.

We'll start by taking the second derivative of the given function, then checking to see at what values of `x` the second derivative is `0`.

`y'=4x^3+3x^2-4x`

`y''=12x^2+6x-4`

To find out where `y''=0`, we will need to use the quadratic formula (warning: these will not be extraordinarily pleasant values).

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-6+-sqrt228)/24`

`x_2=(sqrt(228)-6)/24` and `x_1=(-(6+sqrt(228)))/24`

These values are approximately `x_1=-0.88` and `x_2=0.38`.

We will now pick a few "test points" so that we can check to see if the function is increasing or decreasing around these values of `x`. We want to pick values that lie on either side of our `x`-values. Integers are recommended for simplicity. I would use `-1`,`0`, and `1` in this case. One way to organize this is to draw it out on a number line, shown below.

Now, we put these test points into the second derivative of our function and check to see if they yield negative or positive values. If the value is negative, the function is concave down, and if the value is positive, the function is concave up.

I will write `y` as `f(x)` from this point on.

`f''(-1)` returns a positive value.
`f''(0)` returns a negative value.
`f''(1)` returns a positive value.

Thus, the function is concave up on the interval

(`-∞, (-(6+sqrt(228)))/24`) U (`(sqrt(228)-6)/24,∞`)

And concave down

`((-(6+sqrt(228)))/24, (sqrt(228)-6)/24)`

To find the inflection points, we take the points where the function changes from concave up to concave down and vice versa, and put those `x` values back into the original function to get the corresponding `y` values.

This yields the approximate inflection points (to four decimal places)

`(-0.8792,-0.6279)` and `(0.3792,0.7877)`.