Second Derivative Test for Local Extrema (College Board AP® Calculus AB): Revision Note

Let the function $f$ be defined by $f\left(x\right)=x^3-3x^2+3$.

Find the coordinates of any local extrema on the graph of $f$, and classify the nature of these extrema.

Answer:

Find the critical points by finding the first derivative and setting equal to zero

$f^{\text{'}}\left(x\right)=3x^2-6x$

$\begin{array}{rcl}3x^2-6x& =& 0\\ x^2-2x& =& 0\\ x\left(x-2\right)& =& 0\end{array}$

$x=2$ or $x=0$

Find the $y$-coordinates at these points

$f\left(2\right)=-1$

$f\left(0\right)=3$

Critical points at (2, -1) and (0, 3)

Classify these points by using the second derivative test

$f^{\text{'}\text{'}}\left(x\right)=6x-6$

Substitute the $x$ value of each critical point into the second derivative

$f^{\text{'}\text{'}}\left(2\right)=6$

$f^{\text{'}\text{'}}\left(0\right)=-6$

(2, -1) is a local minimum by the second derivative test because $f\text{'}\left(2\right)=0$ and $f\text{'}\text{'}\left(2\right)>0$

(0, 3) is a local maximum by the second derivative test because $f\text{'}\left(0\right)=0$ and $f\text{'}\text{'}\left(0\right)<0$

Determine the nature of the critical point on the graph of $f\left(x\right)=3x^4+2x^6$.

Answer:

Find the critical point first, by using $f\text{'}\left(x\right)=0$

$f\text{'}\left(x\right)=12x^3+12x^5$

$\begin{array}{rcl}12x^3+12x^5& =& 0\\ 12x^3\left(1+x^2\right)& =& 0\end{array}$

$x=0$ is the only real solution

Find the $y$-coordinate

$f\left(0\right)=0$

The critical point is at (0, 0)

To determine the nature of the critical point, start by finding the second derivative at this point

$f\text{'}\text{'}\left(x\right)=36x^2+60x^4$

$f\text{'}\text{'}\left(0\right)=0$

First derivative is zero, and second derivative is zero

Therefore could be any of minimum, maximum, or point of inflection

Method 1 - Using the first derivative

Check the first derivative either side of the point to determine the nature of the point

$f\text{'}\left(-1\right)=12{\left(-1\right)}^3+12{\left(-1\right)}^5=-24$

$f\text{'}\left(1\right)=12{\left(1\right)}^3+12{\left(1\right)}^5=24$

Just before $x=0$, $f$ is decreasing because $f\text{'}\left(x\right)<0$

Just after $x=0$, $f$ is increasing because $f\text{'}\left(x\right)>0$

Therefore, $f$ has a local minimum at $\left(0,0\right)$ by the first derivative test

Method 2 - Using the second derivative

Check the second derivative either side of the point to determine the nature of the point

$f\text{'}\text{'}\left(-1\right)=36·{\left(-1\right)}^2+60·{\left(-1\right)}^4=96$

$f\text{'}\text{'}\left(1\right)=36·{\left(1\right)}^2+60·{\left(1\right)}^4=96$

Just before $x=0$, $f$ is concave up because $f\text{'}\text{'}\left(x\right)>0$

Just after $x=0$, $f$ is concave up because $f\text{'}\text{'}\left(x\right)>0$

Therefore, $f$ has a local minimum at $\left(0,0\right)$ because $f\text{'}\left(0\right)=0$ and $f$ is concave up around the point