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

\[f'(x) = 6x^2-6x-12,\]

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

\[ \begin{align} f''(-1) &= 12(-1)-6 \\ &= -12-6 \\ &= -18, \end{align}\]

and

\[ \begin{align} f''(2) &= 12(2)-6 \\ &= 24-6 \\ &= 18. \end{align}\]

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

\[g'(x) = 3x^2-6x+3,\]

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

\[ \begin{align} g''(1) &= 6(1)-6 \\ &= 6-6 \\ &= 0. \end{align}\]

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

\[h'(x)=-4x^3,\]

and by using it once again you find

\[h''(x) = -12x^2.\]

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

\[ \begin{align} h''(0) &= -12(0)^2 \\ &= 0. \end{align} \]

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

\[f'(x)=3x^2,\]

and

\[f''(x)=6x.\]

Next, write and solve the inequality \( f''(x) < 0 \) to find where the function is concave, so

\[ \begin{align} f''(x) &< 0 \\ 6x&<0 \\ x &< 0. \end{align} \]

From the above inequality you can conclude that the function is concave in the interval \( (-\infty,0).\) 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

\[ \begin{align} f''(x) &> 0 \\ 6x&>0 \\ x &> 0. \end{align} \]

Therefore, the function is convex in the interval \( (0,\infty).\)

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\[g'(x)=x^2-6x+5.\]
• 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\[g'(c)=c^2-6c+5,\]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 \( g'(x),\) 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

\[ \begin{align} g''(1) &= 2(1)-6 \\ &= -4, \end{align}\]

so there is a local maximum at \(x=1,\) and

\[ \begin{align} g''(5) &= 2(5)-6 \\ &= 4, \end{align}\]

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

\[\begin{align} g(1) &= \frac{1}{3} (1)^3 - 3(1)^2 + 5(1) +4 \\ &= \frac{19}{3} \end{align}\]

is the local maximum and

\[\begin{align} g(5) &= \frac{1}{3} (5)^3 - 3(5)^2 + 5(5) +4 \\ &= -\frac{13}{3} \end{align}\]

is the local minimum.

Next, to find where the function is concave solve the inequality \( g''(x) < 0 \), that is

\[ \begin{align} 2x-6 &< 0 \\ x &< 3, \end{align} \]

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

\[x>3,\]

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

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

\[ f'(x) = x^4-1,\]

and

\[ f''(x) = 4x^3.\]

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

\[ \begin{align} f''(-1) &= 4(-1)^3 \\ &= -4, \end{align} \]

and

\[ \begin{align} f''(1) &= 4(1)^3 \\ &= 4. \end{align} \]

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

\[ g'(x) = 6x^2 -2x +3, \]

and

\[ g''(x) = 12x - 2.\]

Next, write and solve the inequality \( g''(x) < 0 \) to find where the function is concave, that is

\[ \begin{align} g''(x) &< 0 \\ 12x-2&<0 \\ 12x &< 2 \\ x &< \frac{1}{6}, \end{align} \]

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

Finally, write and solve the inequality \( g''(x) > 0 \) to find where the function is convex, that is

\[ \begin{align} g''(x) &> 0 \\ 12x-2&>0 \\ 12x &> 2 \\ x &> \frac{1}{6}. \end{align} \]

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