Second derivative tests are important tools in Calculus used to determine the local extrema and Concavity of a function. The test process involves finding the second derivative functions and evaluating the second derivative value for critical values. The sign or the value of the second derivative is used in describing the behavior of the graphs of the first derivative and hence the function itself.
This page provides you with the details of second derivative tests, their proofs and examples.

A function f is said to be concave upward in an interval I if the graph of the function lies above all the tangents in the interval. The function is said to be concave down if the graph lies below all the tangents in the interval.

The concavity of the function can be determined from the sign of the second derivative in the interval.

 Concave Up     
 Concave Down
                      Concave upward
                       Concave Downward

The second derivative test for concavity is stated as follows:
  1. If f"(x) > 0 for all x in the interval I, then the graph of f(x) is concave upward in the interval.
  2. If f"(x) < 0 for all x in the interval I, then the graph of f(x) is concave downward in the interval.
As a consequence of the concavity test, another application of the second derivative is found for finding the maximum and minimum values of a function in a given interval. It is stated as follows:
If the function f(x) is continuous on the interval (a, b) and f'(c) = 0 for some c ∈ (a, b), then
  1. if f"(c) > 0, then f has a local minimum at c.
  2. if f"(c) < 0, the f has a local maximum at c.
The Second derivative Test for concavity can be proved using Mean Value Theorem.
Given: f(x) is a continuous function in an interval [a, b] and f"(x) > 0 in (a, b)
Prove: f(x) is concave up in the interval (a, b).

Second Derivative Test Proof

Let c be any number in (a,b). To prove f(x) is concave up it has to be shown that the curve y = f(x) lies
above the tangent at the point (c, f(c)).

Since the equation to the tangent line is y = f(c) + f'(c) (x-c) it has to be shown f(x) > f(c) + f'(c) (x-c) for all x ∈ (a,b) and x ≠ c.

Let us first prove this for the case x > c.

Hence by mean value theorem we can find a number 'k' in (c, x) i.e c < k < x such that
f(x) - f(c) = f'(k) (x -c)                       
⇒ f(x) = f(c) + f'(k) (x -c)   ------------------------------(1)
It is given f"(x) > 0 in (a, b) which implies f'(x) is increasing in (a, b)
We have both c and k ∈ (a,b) and c < k
Hence f'(c) < f'(k)
⇒ f'(c) (x -c) < f'(k) (x -c)                          x-c is positive as x > c

Adding f(c) to both the sides of the inequality,
f(c) + f'(c)( x-c) <  f(c) + f'(k) (x-c)
f(c) + f'(c) (x-c) < f(x)                             Substitution using (1)
or
f(x) > f(c) + f'(c) (x -c)
which proves that the curve y = f(x) lies above the tangent line at any point in the given interval.

Case (2) x < c
By mean value theorem, we can find a number 'k' in (x, c) such that
f(c) - f(x) = f'(k) (c-x)
⇒ f(x) = f(c) + f'(k) (x-c)  ---------------------------------------------(2)
As f'(x) is increasing in (a, b)
f'(c) > f'(k)
f'(c) (x-c) < f'(k) (x -c)                                  x-c is negative. The inequality is reversed.

Adding f(c) to both the sides of the inequality,
f(c) + f'(c) (x-c) < f(c) + f'(k) (x-c)
f(c) + f'(c) (x-c) < f(x)                                 Substitution using (2)
or
f(x) > f(c) + f'(c) (x-c)
which is the statement required.

The second derivative test for the case f"(x) < 0 can also be proved, using similar steps.

Solved Example

Question: Let f(x) = x4 - 2x2 + 3
a) Discuss the concavity and find the point of inflection
b) Find the local maximum and minimum values of x
Solution:
 
f(x) = x4 - 2x2 + 3
     f'(x) = 4x3 - 4x                                                 First derivative
     f"(x) = 12x2 -4                                                 Second derivative
     We can find the critical numbers equating f'(x) = 0
     4x3 - 4x = 0    ⇒    4x (x +1) (x-1) = 0               Factored form
      thus the critical numbers are x = 0, -1 and 1
      Let us check the second derivative value at these critical numbers.
      f"(0) = 12(0) - 4 = -4    f"(0) is negative ⇒ x = 0 is a maximum point
      f"(-1) = 12(-1)2 - 4 = 12 - 4 = 8.  f"(-1) is positive ⇒ x = -1 is a minimum point.
      f"(1) = 12(1)2 - 4 = 12 - 4 = 8. f"(1) is positive ⇒ x = 1 is a minimum point.
      Let us now check the concavity in the intervals formed by the critical points.
     
Interval f"(x) = 12x2 - 4
   Concavity
 (-∞, -1)     Positive  Concave up
 (-1, 0)
    Negative
 Concave down
 (0, 1)     Negative
 Concave down
 (1, ∞)    Positive
 Concave up

To find the inflection point f"(x) = 0
                                         12x2 - 4 = 0

                                          x = ± $\frac{\sqrt{3}}{3}$ ≈ ± 0.577

Now we check the concavity around these points
f"(-0.6) = 0.32 positive and f"(-0.5) = -1 negative
Hence - 0.577 is a point of inflection as the concavity changes from up to downward.
Same way f"(0.5) is negative and f"(0.6) is positive.
x = 0.577 is also a point of inflection as the concavity changes from down to up.

Let us look at the graph of f(x) = x4 - 2x2 + 3

Second Derivative Example

The graph confirms the features found using Second Derivative Test.
 

Second Derivative Test is also used in Multivariable Calculus to determine the local extrema of functions of two variables.

For this purpose the second partial derivatives of the function are found and a discriminant D is defined.

Suppose a function f(x,y) and its first and second partial derivatives are continuous in a disk with center (a,b) and suppose
fx (a,b) = 0 and fy (a,b) = 0 [ (a,b) is a critical point].
Let D = D(a,b) = fxx (a,b) .fyy (a,b) - [fxy (a,b)]2.
1.If D > 0 and fxx (a,b) > 0, then f(a,b) is a local minimum.
2.If D > 0 and fxx (a,b) < 0, then f(a,b) is a local maximum.
3.If D < 0, f(a.b) is not an extremum. The point (a.b) is then called a saddle point.