Question

How do you use the second derivative test to find the relative maxima and minima of the given `f(x)= x^4 - (2x^2) + 3`?

Answer 1

A relative maximum is where the first derivative is null and the second derivative is negative.
A relative minimum is where the first derivative is null and the second derivative is positive.

Explanation:

`f(x) = x^4-(2x^2)+3` gives you a curve.

We want the curve's relative maxima and minima, that is, where the curve stops increasing to start decreasing or vice-versa. At these points, the curve's slope will be null.

`(df)/dx=4x^3-4x` gives you the variations of this curve's slope.

Let's look for the solutions to `4x^3-4x=0`

`4x^3-4x=(4x^2-4)x=0`
We have three solutions here: `x_0=0;x_+=1;x_(-)=-1`

Now we want to know if these points are maxima or minima.

If a point is a maximum , the curve will be increasing before reaching the point and be decreasing after passing the point.

If a point is a minimum , the curve will be decreasing before reaching the point and be increasing after passing the point.

When a curve is increasing, its slope is positive.
When a curve is decreasing, its slope is negative.

So we want to know if, at a given point, the slope (first derivative) is:

negative-null-positive `rarr` minimum
or
positive-null-negative `rarr` maximum

To do so, we use the second derivative:

`(d^2f)/dx^2=12x^2-4` with the `x_0;x_+;x_(-)` points:

`12*0^2-4=-4 rarr` the slope is decreasing around 0, therefore we are in a "positive-null-negative" situation, therefore, we have a maximum here.

`12*(-1)^2-4=12-4=8 rarr` the slope is increasing around 0, therefore we are in a "negative-null-positive" situation, therefore, we have a minimum here.

`12*1^2-4=12-4=8 rarr` minimum.