The definition of the **concavity** of a graph is introduced along with inflection points. Examples, with detailed solutions, are used to clarify the concept of concavity.

Figure 1

Figure 2

Let \( f' \) be the first derivative of function \( f \) that is differentiable on a given interval \( I \), the graph of \( f \) is

(i) **concave up** on the interval \( I \), if \( f' \) is increasing on \( I \),

or

(ii) **concave down** on the interval \( I \), if \( f' \) is decreasing on \( I \).

The sign of the second derivative informs us when \( f' \) is increasing or decreasing.

Let \( f'' \) be the second derivative of function \( f \) on a given interval \( I \), the graph of \( f \) is

(i) **concave up** on \( I \) if \( f''(x) > 0 \) on the interval \( I \).

(ii) **concave down** on \( I \) if \( f''(x) < 0 \) on the interval \( I \).

A point \( P \) on the graph of \( y = f(x) \) is a point of inflection if \( f \) is continuous at \( P \) and the concavity of the graph changes at \( P \). In view of the above theorem, there is a point of inflection whenever the second derivative changes sign.

\( f'(x) = 2ax + b \)

\( f''(x) = 2a \)

We now study the sign of \( f''(x) \) which is equal to \( 2a \). If \( a \) is positive, \( f''(x) \) is positive in the interval \((-8 , + 8)\). According to the theorem above, the graph of \( f \) will be concave up for positive values of \( a \).

If \( a \) is negative, the graph of \( f \) will be concave down on the interval \((-8 , + 8)\) since \( f''(x) = 2a \) is negative.

The graphs of two quadratic functions are shown below: \( y = 2x^2 - 2x - 1 \) whose graph is concave up because its leading coefficient (\( a = 2 \)) is positive and \( y = -x^2 + 3x + 1 \) whose graph is concave down because its leading coefficient (\( a = -1 \)) is negative.

b) Use a graphing calculator to graph \( f \) and confirm your answers to part a).

a)

\( f'(x) = 4x^3 - 6x^2 + 1 \)

\( f''(x) = 12x^2 - 12x \)

Find the zeros \( f''(x) \).

\( 12x^2 - 12x = 0 \)

\( 12x(x - 1) = 0 \)

Two zeros

\( x = 0 \) and \( x = 1 \)

Study sign of \( f'' \)

The two zeros split the set of real numbers into three intervals. Select a value for \( x \) in each of the three intervals and find the sign of \( f'' \)

We now use the table of sign and the theorem above to conclude that

in the interval \((-8 , 0)\); \( f'' \) is positive and therefore the graph of \( f \) is concave up

in the interval \( (0 , 1) \); \( f'' \) is negative and therefore the graph of \( f \) is concave down

in the interval \( (1 , +8) \); \( f'' \) is positive and therefore the graph of \( f \) is concave up

The second derivative \( f'' \) changes sign at \( x = 0 \) and \( x = 1 \) and therefore the graph of \( f \) has two inflection point: \( (0 , f(0)) \) and \( (1 , f(1)) \)

b)

The graph of \( f \) (blue) and \( f'' \) (red) are shown below. It can easily be seen that whenever \( f'' \) is negative (its graph is below the x-axis), the graph of \( f \) is concave down and whenever \( f'' \) is positive (its graph is above the x-axis) the graph of \( f \) is concave up.

Point \( (0,0) \) is a point of inflection where the concavity changes from up to down as \( x \) increases (from left to right) and point \( (1,0) \) is also a point of inflection where the concavity changes from down to up as \( x \) increases (from left to right).

a)

On the interval \((-\infty , 2)\), the graph of \( f'' \) is below the x-axis and therefore \( f'' \) is negative, hence \( f \) is concave down on the interval \((-\infty , 2)\)

On the interval \( (2 , +\infty) \), the graph of \( f'' \) is above the x-axis and therefore \( f'' \) is positive, hence \( f \) is concave up on the interval \( (2 , +\infty) \)

At \( x = 2 \), the sign of \( f'' \) changes and therefore \( x = 2 \) is a point of inflection.

a)

On the interval \((-\infty , -2)\), \( f' \) decreases and therefore \( f'' \) is negative; the graph of \( f \) is concave down

On the interval \( (-2 , -1) \), \( f' \) increases and therefore \( f'' \) is positive; the graph of \( f \) is concave up

On the interval \( (-1 , 1) \), \( f' \) decreases and therefore \( f'' \) is negative; the graph of \( f \) is concave down

On the interval \( (1 , +\infty) \), \( f' \) increases and therefore \( f'' \) is positive; the graph of \( f \) is concave up

The concavity of the graph of \( f \) changes at \( x = -2 \), \( x = -1 \) and \( x = 1 \) and therefore these are all point of inflection.

Calculus Tutorials and Problems