# 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?

Jun 28, 2015

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:

$f \left(x\right) = {x}^{4} - \left(2 {x}^{2}\right) + 3$ 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.

$\frac{\mathrm{df}}{\mathrm{dx}} = 4 {x}^{3} - 4 x$ gives you the variations of this curve's slope.

Let's look for the solutions to $4 {x}^{3} - 4 x = 0$

$4 {x}^{3} - 4 x = \left(4 {x}^{2} - 4\right) x = 0$
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 $\rightarrow$ minimum
or
positive-null-negative $\rightarrow$ maximum

To do so, we use the second derivative:

$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = 12 {x}^{2} - 4$ with the x_0;x_+;x_(-) points:

$12 \cdot {0}^{2} - 4 = - 4 \rightarrow$ the slope is decreasing around 0, therefore we are in a "positive-null-negative" situation, therefore, we have a maximum here.

$12 \cdot {\left(- 1\right)}^{2} - 4 = 12 - 4 = 8 \rightarrow$ the slope is increasing around 0, therefore we are in a "negative-null-positive" situation, therefore, we have a minimum here.

$12 \cdot {1}^{2} - 4 = 12 - 4 = 8 \rightarrow$ minimum.