


The Sign of the Second DerivativeSummaryThe 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.
ExampleConsider the graph of f(x) = 3x^{4} − 8x^{3} − 90x^{2}. First obtain the first and second derivatives. f '(x) = 12x^{3} − 24x^{2} − 180x and f "(x) = 36x^{2} − 48x − 180. To find any stationary points, solve the equation f '(x) = 0. f '(x) = 0 = 12x^{3} − 24x^{2} − 180x = 12x(x^{2} − 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) = 36x^{2} − 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 ycoordinate of the point, calculate the value of the function for each value of x. f(x) = 3x^{4} − 8x^{3} − 90x^{2} 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 = 36x^{2} − 48x − 180 = 12(3x^{2} − 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 ycoordinate of these points of inflection, calculate the value of the function for each value of x. f(x) = 3x^{4} − 8x^{3} − 90x^{2} so f() = 189 and f(3) = 783.
f "(x) = 12(3x + 5)(x − 3) ExampleConsider the graph of f(x) = . First obtain the first and second derivatives. Since , 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.
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