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Second Derivative Test

- Calculus
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Calculus is a powerful tool for identifying the properties and the shape of a graph. One example is The First Derivative Test used to find the critical points of a function. These points tell you where the slope of a graph is 0.

Sometimes these critical points are local maxima or minima, and sometimes neither. How can you determine if the critical points are local maxima or minima? Use The Second Derivative Test!

For a review on maximums and minimums see our articles Maxima and Minima and Absolute Maxima and Minima.

After finding the critical points of a function through The First Derivative Test, the next question that arises naturally is if those points are local maxima, local minima, or neither. **The Second Derivative Test **is a **method** for telling whether a critical point is a local extremum.

Let \( f(x) \) be a function that can be differentiated at least twice, and let \( c \) be a critical point of \( f(x).\) **The Second Derivative Test **states that:

- If \( f''(c) < 0,\) then \( f(c) \) is a local maximum of \( f(x). \)
- If \( f''(c) > 0,\) then \( f(c) \) is a local minimum of \( f(x). \)

In words, The Second Derivative Test tells you the following:

If the second derivative at a critical point is negative, the function has a local maximum at that point.

If the second derivative at a critical point is positive, the function has a local minimum at that point.

Note that the Second Derivative Test does not cover the case when \( f''(c)=0. \) That is because the test becomes **inconclusive **in that scenario, so there is nothing more you can tell about the function.

The Second Derivative Test receives its name from the fact that you need to find the second derivative of the function you are working with. Makes sense, right?

Since the Second Derivative Test is capable of telling you whether a critical point is a local maximum or a local minimum, it is usually used after finding the critical points using The First 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 \( x=2. \)

It is worth noting that The Second Derivative Test only tells us what happens if the second derivative is NOT equal to zero. Recall that if \( f''(c)=0 \) the test becomes** inconclusive. **

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.

Note that this function **does not **have relative extrema.

In the above example, the function did not have any relative extrema. Remember that you could not conclude anything from The Second Derivative Test because \( f''(c)=0,\) but that does not mean that this implies that a function will never 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.

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!

Just like the first derivative tells you whether a function is increasing or decreasing, the second derivative tells us about the **concavity **of the graph.

A function is said to be **concave down**, or just **concave**, in an interval where its second derivative is negative. The lines tangent to the function's graph inside an interval where it is concave will lie **above **the graph.

If a critical point is inside an interval where the function is concave, then it will be a maximum because of The Second Derivative Test. The following graph shows lines tangent to a function in an interval where it is concave.

If the second derivative of a function in a given interval is positive instead, the function is said to be **concave up** or **convex.**

A function is said to be **concave up,** or **convex**, in an interval where its second derivative is positive. The lines tangent to the function's graph inside an interval where it is convex will lie **below **the graph.

Like with a concave interval, if a critical point is inside an interval where the function is convex, it will be a minimum because of The Second Derivative Test. The following graph shows lines tangent to a function in an interval where it is convex.

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 \( (0,\infty).\)

The x-values for where the concavity of a graph change are called **inflection points. **If a function's second derivative is continuous then the second derivative evaluated at an inflection point is equal to 0.

Let \( f(x) \) be a function that:

- Is continuous at \(x=a.\)
- Changes its concavity at \(x=a.\)

The \( a \) value is called an **inflection point** of \( f.\)

The point \( x=0 \) from the previous example is an inflection point.

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

This means that the equation can be factored as\[ (c-1)(c-5)=0, \] which means that the solutions are \(c=1\) and \(c=5.\)Number 1 Number 2 Product Sum 1 5 5 6 -1 -5 5 -6

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**.

Just like The First Derivative Test, when doing The Second Derivative Test for multivariable functions you need to use Partial Derivatives. However, this test is more complex when evaluating multivariable functions, and it is out of the scope of this article.

We will now take a look at more examples of The Second Derivative Test.

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 \(x=1.\)

We will now use the second derivative of a function to inspect its concavity.

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 \( (\frac{1}{6},\infty). \)

- The Second Derivative Test is a method for telling what kind of extremum is a critical point.
- If the second derivative at a critical point is
**negative**, the function has a local**maximum**at that point. - If the second derivative at a critical point is
**positive**, the function has a local**minimum**at that point.

- If the second derivative at a critical point is
- In case that the second derivative at a critical point is equal to
**zero**the test is**inconclusive**. This means that the critical point can be a local maximum, a local minimum, or neither. - The second derivative of a function also gives information about the shape of a graph.
- The graph is concave in an interval where its second derivative is negative.
- The graph is convex in an interval where its second derivative is positive.

After finding the critical points, or stationary points, of a function you evaluate the second derivative at those points.

- If the second derivative at a critical point is negative, the function has a local maximum at that point.
- If the second derivative at a critical point is positive, the function has a local minimum at that point.
- If the second derivative at a critical point is equal to zero, the test becomes inconclusive.

Now take the cosine function, its derivative is the negative sine function. Since you differentiated twice you can say that the second derivative of the sine function is the negative sine function.

More about Second Derivative Test

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