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The Sign of the Second Derivative

Summary

The second derivative tells us a lot about the qualitative behaviour of the graph of a function. This is especially important at points close to the critical (stationary) points. Critical points occur where the first derivative is 0.

1. If the second derivative is positive at a point, the graph is concave up at that point. If the second derivative is positive at a critical point, then the critical point is a local minimum.
2. If the second derivative is negative at a point, the graph is concave down. If the second derivative is negative at a critical point, then the critical point is a local maximum.
3. An inflection point marks the transition from concave up to concave down or vice versa. The second derivative will be zero at an inflection point.

Example

Consider the graph of f(x) = 3x⁴ − 8x³ − 90x².

First obtain the first and second derivatives.

f '(x) = 12x³ − 24x² − 180x and f "(x) = 36x² − 48x − 180.

To find any stationary points, solve the equation f '(x) = 0.

f '(x) = 0 = 12x³ − 24x² − 180x = 12x(x² − 2x − 15) = 12x(x + 3)(x − 5)

There are three solutions, x = 0, x = -3 and x = 5.

To determine the type of stationary point, calculate the second derivative at each value of x.

f "(x) = 36x² − 48x − 180 so f "(0) = -180, f "(-3) = 288 and f "(5) = 480.

We have respectively a local maximum, then two local minima.

To determine the y-coordinate of the point, calculate the value of the function for each value of x.

f(x) = 3x⁴ − 8x³ − 90x² so f(0) = 0, f(-3) = -351 and f(5) = -1375.

Hence (0,0) is a local maximum, (-3,-351) is a local minimum and (5,-1375) is a local minimum.

To determine any points of inflection, solve the equation f "(x) = 0, and determine the concavity of the graph.

f "(x) = 0 = 36x² − 48x − 180 = 12(3x² − 4x − 15) = 12(3x + 5)(x − 3).

There are two solutions, x = - and x = 3 .

If x < - then f "(x) is positive and the graph is concave upward. When - < x < 3 f "(x) is negative and the graph is concave downward. Finally when x > 3 , f "(x) is positive and the graph is concave upward. Because the concavity changes at x = - and x = 3 both give points of inflection.

To determine the y-coordinate of these points of inflection, calculate the value of the function for each value of x.

f(x) = 3x⁴ − 8x³ − 90x² so f(-) = -189 and f(3) = -783.

f "(x) = 12(3x + 5)(x − 3)

Example

Consider the graph of f(x) = .

First obtain the first and second derivatives.

Since , f '(x) =
and using the Product Rule,
f "(x) = .

To find any stationary points, solve the equation f '(x) = 0.

f '(x) = 0 = has just the solution x = 0.

f "(x) = so f "(0) = and, being positive, indicates a local minimum.

f(0) = , so the minimum is at (0,).

To determine any points of inflection, solve the equation f "(x) = 0.

f "(x) = 0 = has no real solutions, so there are no points of inflection.

To determine the concavity of the graph, consider the sign of the second derivative.

f "(x) = has a numerator that is always positive. The denominator is positive if |x| < 3 but negative if |x| > 3. Hence f "(x) is negative and the curve concave downwards when x < -3 and x > 3. f "(x) is positive and the curve concave upwards when -3 < x < 3.

If we note that the function has no defined value when x = -3 and x = 3, we can sketch its graph.

Exercise

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