# Aim: Concavity & 2 nd Derivative Course: Calculus Do Now: Aim: The Scoop, the lump and the Second Derivative. Find the critical points for f(x) = sinxcosx;

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Aim: Concavity & 2 nd Derivative Course: Calculus Do Now: Aim: The Scoop, the lump and the Second Derivative. Find the critical points for f(x) = sinxcosx; [0, 2  ]

Aim: Concavity & 2 nd Derivative Course: Calculus Definition of Concavity Let f be differentiable on an open interval I. The graph of f is concave upward on I if f’ is increasing on the interval and concave downward on I if f’ is decreasing on the interval. Concave upward: the graph of f at (c, f(c)) lies above the tangent line.

Aim: Concavity & 2 nd Derivative Course: Calculus Definition of Concavity Let f be differentiable on an open interval I. The graph of f is concave upward on I if f’ is increasing on the interval and concave downward on I if f’ is decreasing on the interval. Concave downward: the graph of f at (c, f(c)) lies below the tangent line.

Aim: Concavity & 2 nd Derivative Course: Calculus Definition of Concavity Let f be differentiable on an open interval I. The graph of f is concave upward on I if f’ is increasing on the interval and concave downward on I if f’ is decreasing on the interval. Concave down Concave up slope m = 0 slope m = -1 f f’

Aim: Concavity & 2 nd Derivative Course: Calculus Test for Concavity Using 2 nd Derivative Set f be a function whose second derivative exists on an open Interval I. 1.If f”(x) > 0 in I, then the graph of f is concave upward in I. 2. If f”(x) < 0 for all x in I, then the graph of f is concave downward in I. Note: If f”(x) = 0 for all x in I, then f is linear.

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem 1 Determine the open intervals on which the graph of f(x) = 6(x 2 + 3) -1 is concave upward or downward. 1. Find second derivative 1 st derivative 2 nd derivative

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem 1 Determine the open intervals on which the graph of f(x) = 6(x 2 + 3) -1 is concave upward or downward. 2. Solve f’’(x) = 0 x = ± 1 f’’(x) is defined for all reals

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem 1 Determine the open intervals on which the graph of f(x) = 6(x 2 + 3) -1 is concave upward or downward. 3. Test the intervals Interval Test Value x = -2x = 0x = 2 Sign of f”(x) f’’(-2) > 0f’’(0) < 0f’’(2) > 0 Conclusion Concave up Concave down Concave up Concave down Concave up f”(x) > 0 f”(x) < 0

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem 2 Determine the open intervals on which the graph of is concave upward or downward. 1. Find second derivative 1 st derivative

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem 2 Determine the open intervals on which the graph of is concave upward or downward. 1. Find second derivative (con’t) 1 st derivative 2 nd derivative

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem 2 Determine the open intervals on which the graph of is concave upward or downward. 2. Solve f’’(x) = 0 undefined @ x = ±2 no real values of x result in 0

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem 2 Determine the open intervals on which the graph of is concave upward or downward. 3. Test the intervals Interval Test Value x = -3x = 0x = 3 Sign of f”(x) f’’(-3) > 0f’’(0) < 0f’’(3) > 0 Conclusion Concave up Concave down Concave up Concave down Concave up f”(x) > 0 f”(x) < 0 f”(x) > 0

Aim: Concavity & 2 nd Derivative Course: Calculus Do Now: Aim: The Scoop, the lump and the Second Derivative. Use the 2 nd derivative to determine the open intervals on which the graph of is concave up and concave down.

Aim: Concavity & 2 nd Derivative Course: Calculus Points of Inflection A differentiable point on a graph where concavity changes. Concave up Concave down Concave up points of inflection f”(x) = 0

Aim: Concavity & 2 nd Derivative Course: Calculus Points of Inflection A differentiable point on a graph where concavity changes. In other words where the derivative stops increasing and starts decreasing or vice versa. Concave up Concave down Concave up Concave down Note: graph of tangent crosses f at point of inflection

Aim: Concavity & 2 nd Derivative Course: Calculus Points of Inflection To locate possible points of inflection, determine the values of x for which f’’(x) = 0 or for which f’’ is undefined. If (c), f(c)) is point of inflection of the graph of f, then either f’’(c) = 0 or f”(c) is undefined at x = c. Concave up Concave down (c), f(c)) f’’(c) = 0

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem 3 Determine the points of inflection and discuss the concavity of the graph of f(x) = x 4 – 4x 3. 1. Find second derivative 1 st derivative 2 nd derivative

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem 3 Determine the points of inflection and discuss the concavity of the graph of f(x) = x 4 – 4x 3. 2. Solve f’’(x) = 0 x = 0, 2 are possible points of inflection

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem 3 Determine the points of inflection and discuss the concavity of the graph of f(x) = x 4 – 4x 3. 3. Test Intervals to verify concavity Interval Test Value x = -1x = 1x = 3 Sign of f”(x) f’’(-1) > 0f’’(1) < 0f’’(3) > 0 Conclusion Concave up Concave down Concave up

Aim: Concavity & 2 nd Derivative Course: Calculus Point of Inflection f’’(x) = 0 is not always a point of inflection f’’(x) = 12x 2 f’’(0) = 12(0) 2 = 0 both intervals: are concave upwards, i.e. no change of concavity

Aim: Concavity & 2 nd Derivative Course: Calculus Second Derivative Test The second derivative can also be used to test for relative maxima and minima. c Concave up If f’(c) = 0 and f’’(c) > 0, then f(c) is a relative minimum. c Concave down If f’(c) = 0 and f’’(c) < 0, then f(c) is a relative maximum.

Aim: Concavity & 2 nd Derivative Course: Calculus Second Derivative Test The second derivative can also be used to test for relative maxima and minima. Let f be a function such that f’(c) = 0 and the second derivative of f exists on an open interval containing c. 1.If f’’(c) > 0, then f(c) is a relative minimum. 2.If f’’(c) < 0, then f(c) is a relative maximum. If f’’(c) = 0, the test fails. In such cases, you can use the First Derivative Test

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem Find the relative extrema for f(x) = -3x 5 + 5x 3. 1. Find the relative extrema 1 st derivative 2 nd derivative x = -1, 0, 1 2. Second Derivative Test – find 2 nd (-1, -2), (0, 0), (1, 2)

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem Find the relative extrema for f(x) = -3x 5 + 5x 3. 3. Evaluate 2 nd Derivative for extrema PointSign of f’’Conclusion f’’(-1) = 30 > 0 Relative minimum f’’(1) = -30 < 0 Relative maximum f’’(0) = 0Test fails (-1, -2) (1, 2), (0, 0) What can you interpret from this failure?

Aim: Concavity & 2 nd Derivative Course: Calculus Model Problem Find the relative extrema for f(x) = -3x 5 + 5x 3. f’’(0) = 0Test fails (0, 0) Using 1 st derivative test you’ll see that x increases to left and right of x = 0. a point of inflection we already knew the slope of tangent at the point (0, 0) was 0. (-1, -2) (1, 2) (0, 0) What can you interpret from this failure?

Aim: Concavity & 2 nd Derivative Course: Calculus The 2 nd Derivative & Points of Inflection f g h u c If a critical point has a zero first derivative (horizontal tangent), then there is a high point if the 2 nd derivative is negative (concave down) or a low point if the second derivative is positive (concave up).

Aim: Concavity & 2 nd Derivative Course: Calculus Critical, End, Inflection Points etc. concavity up down e.p. max. p.i. cusp max.p.i. p.i. cusp min. p.i. plateau max. cusp no p.i. min. p.i. e.p. min. abcdefg p.i. – point of inflection cusp – sharp corner e.p. – end point

Aim: Concavity & 2 nd Derivative Course: Calculus 2 nd Derivative Test Failures When f’(c) = 0 and f”(c) = 0 2 nd Derivative tests fails – Use the 1 st Derivative test instead.

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