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## 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

 Calculate the first and second derivative of x4 + x3 + x2 + x + when x = Working: f'( ) =    f"( ) = The graph of f(x) close to x = will be concave down concave up point of inflection local maximum local minimum
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