Question

How do you use the second derivative test to find the relative maxima and minima of the given $f\left( x \right)={{x}^{4}}-\left( 2{{x}^{2}} \right)+3$

Answer 1

We are said to find the relative maxima and minima using the second derivative test. So, we will start with finding the first derivative and then equal that to zero to find the points of maxima or minima. Again, we will find the second derivative to find that in which point we have maximum or in which point minimum values.

Complete step by step solution:
According to the question, we have, $f\left( x \right)={{x}^{4}}-\left( 2{{x}^{2}} \right)+3$ gives you a curve.
Here we want to find 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.
Now, to start with, we will find, $\dfrac{df}{dx}$ which gives you the variations of this curve's slope.
$\dfrac{df}{dx}=4{{x}^{3}}-4x$
Let's look for the solutions to $\dfrac{df}{dx}=0$
Thus we get, $\dfrac{df}{dx}=4{{x}^{3}}-4x=0$
$\Rightarrow 4x\left( {{x}^{2}}-1 \right)=0$
We are getting, 4x = 0 or ${{x}^{2}}-1=0$
So, from the first equation, x = 0, and second equation gives us, ${{x}^{2}}=1\Rightarrow x=\pm 1$
We have three solutions here: x = 0 ,1, -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 positive gives us the minimum value of the function and negative slope gives us the maximum value of the function.
To do so, we use the second derivative now.
$\dfrac{{{d}^{2}}f}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{df}{dx} \right)$
Thus we get, $\dfrac{d}{dx}\left( \dfrac{df}{dx} \right)=\dfrac{d}{dx}\left( 4{{x}^{3}}-4x \right)=12{{x}^{2}}-4$
Now, putting the values,
For x = 0, $\dfrac{{{d}^{2}}f}{d{{x}^{2}}}=12.{{\left( 0 \right)}^{2}}-4=0-4=-4$
The slope is decreasing around 0, therefore, we have a maximum here.
For x = 1, $\dfrac{{{d}^{2}}f}{d{{x}^{2}}}=12.{{\left( 1 \right)}^{2}}-4=12-4=8$
The slope is increasing around 1, therefore, we have a minimum here.
For x = -1, $\dfrac{{{d}^{2}}f}{d{{x}^{2}}}=12.{{\left( -1 \right)}^{2}}-4=12-4=8$
The slope is increasing around - 1, therefore, we have a minimum here.

Note: In this problem, we are trying to find the relative maximum and minimum of a function. 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.