Second Derivative Test

Consider the function

$$f(x)=2x^3-3x^2-12x+4,$$

whose critical points are at $x=-1$ and $x=2.$ Use the Second Derivative Test to find whether each critical point is a local maximum or a local minimum.

Answer:

You need the second derivative of the function, so you need to differentiate it twice. Use the Power Rule to find the first derivative, that is

$$f^{\prime }(x)=6x^2-6x-12,$$

and use it again to find the second derivative, so

$$f^{″}(x)=12x-6.$$

Next, you need to evaluate the second derivative at each critical point. Since you are already given the critical points this becomes straightforward, so

$$\begin{array}{rl}{f}^{″}(-1)& =12(-1)-6\\ & =-12-6\\ & =-18,\end{array}$$

and

$$\begin{array}{rl}{f}^{″}(2)& =12(2)-6\\ & =24-6\\ & =18.\end{array}$$

You just found that $f^{″}(-1)=-18,$ which is less than 0. That means there is a relative maximum at $x=-1.$ Likewise, since you found that $f^{″}(2)=18,$ which is greater than 0, you can conclude that a relative minimum occurs at $x=2.$

Fig. 2. Graph of a cubical polynomial and its local extrema

Consider the function

$$g(x)=x^3-3x^2+3x-2,$$

which only has one critical point at $x=1.$ Use the Second Derivative Test to tell if it is a local maximum or a local minimum.

Answer:

You need the second derivative of the function. Start by finding its first derivative using the Power Rule, that is

$$g^{\prime }(x)=3x^2-6x+3,$$

and use it once again to find the second derivative, so

$$g^{″}(x)=6x-6.$$

Next, evaluate the second derivative at its critical point, which is $x=1,$ to obtain

$$\begin{array}{rl}{g}^{″}(1)& =6(1)-6\\ & =6-6\\ & =0.\end{array}$$

Since you evaluated the second derivative at a critical point and got 0, the second derivative test is inconclusive, which means that it cannot tell by itself if there is a local maximum, minimum, or neither.

Now take a look at the graph of the function.

Fig. 3. Graph of a cubic polynomial with no relative extrema

Note that this function does not have relative extrema.

Consider the function

$$h(x)=1-x^4,$$

which only has one critical point at $x=0.$ Use the Second Derivative Test to tell if it is a local maximum or a local minimum.

Answer:

As usual, begin by finding the required derivatives. By using the Power Rule you can find that

$$h^{\prime }(x)=-4x^3,$$

and by using it once again you find

$$h^{″}(x)=-12x^2.$$

Next, evaluate the second derivative at the critical point, so

$$\begin{array}{rl}{h}^{″}(0)& =-12(0{)}^2\\ & =0.\end{array}$$

Once again, since $h^{″}(0)=0$ you cannot conclude anything from The Second Derivative Test. This function, however, has a local maximum located at the critical point, as shown in its graph.

Fig. 4. Graph of a quartic polynomial with a local maximum

In other words, you can't conclude that a point isn't a minimum or maximum just because the Second Derivative Test failed to give you an answer!

Find the intervals for which the function

$$f(x)=x^3+1$$

is concave and convex.

Answer:

To find the intervals where the given function is concave or convex you need to inspect its second derivative, so use the Power Rule twice to find it, that is

$$f^{\prime }(x)=3x^2,$$

and

$$f^{″}(x)=6x.$$

Next, write and solve the inequality $f^{″}(x)<0$ to find where the function is concave, so

$$\begin{array}{rl}{f}^{″}(x)& <0\\ 6x& <0\\ x& <0.\end{array}$$

From the above inequality you can conclude that the function is concave in the interval $(-\mathrm{\infty },0).$ This means that for all x-values less than 0, the function is concave.

To find where the function is convex you write and solve the inequality $f^{″}(x)>0,$ so

$$\begin{array}{rl}{f}^{″}(x)& >0\\ 6x& >0\\ x& >0.\end{array}$$

Therefore, the function is convex in the interval $(0,\mathrm{\infty }).$

Consider the function

$$g(x)=\frac{1}{3}{x}^3-3x^2+5x+4.$$

State its local extrema, inflection points, and concave/convex intervals, if any.

Answer:

You can use the First Derivative Test to find its local extrema.

• Find the derivative of the function.Using the Power Rule to differentiate the given function will give you$$g^{\prime }(x)=x^2-6x+5.$$
• Evaluate the function at a critical point $c$ and set it equal to 0.This step is rather straightforward, evaluate the derivative at $x=c,$ that is$$g^{\prime }(c)=c^2-6c+5,$$and set it equal to 0, so$$c^2-6c+5=0.$$
• Solve the obtained equation for $c.$The above quadratic equation can be solved by factoring. Think of two numbers whose product is $5$ and their sum is $-6.$ Usually, you should start factoring, so

  Number 1	Number 2	Product	Sum
  1	5	5	6
  -1	-5	5	-6

  This means that the equation can be factored as$$(c-1)(c-5)=0,$$ which means that the solutions are $c=1$ and $c=5.$

Now that you found the critical points, you need to find the second derivative. You can do this by using the Power Rule again on $g^{\prime }(x),$ so

$$g^{″}(x)=2x-6.$$

Next, evaluate the second derivative at both critical points to tell whether they are a maxima or minima, that is

$$\begin{array}{rl}{g}^{″}(1)& =2(1)-6\\ & =-4,\end{array}$$

so there is a local maximum at $x=1,$ and

$$\begin{array}{rl}{g}^{″}(5)& =2(5)-6\\ & =4,\end{array}$$

so there is a local minimum at $x=5.$

The maximum and minimum values can be found by evaluating the function at those spots, so

$$\begin{array}{rl}g(1)& =\frac{1}{3}(1{)}^3-3(1{)}^2+5(1)+4\\ & =\frac{19}{3}\end{array}$$

is the local maximum and

$$\begin{array}{rl}g(5)& =\frac{1}{3}(5{)}^3-3(5{)}^2+5(5)+4\\ & =-\frac{13}{3}\end{array}$$

is the local minimum.

Next, to find where the function is concave solve the inequality $g^{″}(x)<0$, that is

$$\begin{array}{rl}2x-6& <0\\ x& <3,\end{array}$$

so the function is concave in the interval $(-\mathrm{\infty },3).$ To find where it is convex just flip the inequality, so

$$x>3,$$

which means that the function is convex in the interval $(3,\mathrm{\infty }).$

Since the function switches from concave to convex at $x=3,$ this is an inflection point.

Fig. 8. A function with its local extrema, inflection point, and concave/convex regions

Consider the function

$$f(x)=\frac{1}{5}{x}^5-x,$$

whose critical points are at $x=-1$ and $x=1.$ Use the Second Derivative Test to find whether each critical point is a local maximum or a local minimum.

Answer:

Use the Power Rule twice to find the second derivative of $f(x),$ that is

$$f^{\prime }(x)=x^4-1,$$

and

$$f^{″}(x)=4x^3.$$

Next, evaluate the second derivative at each critical point, so

$$\begin{array}{rl}{f}^{″}(-1)& =4(-1{)}^3\\ & =-4,\end{array}$$

and

$$\begin{array}{rl}{f}^{″}(1)& =4(1{)}^3\\ & =4.\end{array}$$

Since you found that $f^{″}(-1)<0$ there is a local maximum located at $x=-1.$ Also, since you found that $f^{″}(1)>0$ there is a local minimum located at $x=1.$

Find the intervals for which the function

$$g(x)=2x^3-x^2+3x+1$$

is concave and convex.

Answer:

Begin by using the Power Rule twice to find the second derivative of $g(x),$ so

$$g^{\prime }(x)=6x^2-2x+3,$$

and

$$g^{″}(x)=12x-2.$$

Next, write and solve the inequality $g^{″}(x)<0$ to find where the function is concave, that is

$$\begin{array}{rl}{g}^{″}(x)& <0\\ 12x-2& <0\\ 12x& <2\\ x& <\frac{1}{6},\end{array}$$

so the function is concave in the interval $(-\mathrm{\infty },\frac{1}{6}).$

Finally, write and solve the inequality $g^{″}(x)>0$ to find where the function is convex, that is

$$\begin{array}{rl}{g}^{″}(x)& >0\\ 12x-2& >0\\ 12x& >2\\ x& >\frac{1}{6}.\end{array}$$

Therefore, the function is convex in the interval $(\frac{1}{6},\mathrm{\infty }).$