Ch. 5 – Applications of Derivatives 5.3 – Connecting f’ and f’’ with the Graph of f

Concavity: The curvature of the graph at a point A graph is concave up at x = c if it curves upwards (slope is increasing) at x = c Concave up means f’(c) is increasing and f’’(c) > 0 ! A graph is concave down at x = c if it curves downwards (slope is decreasing) at x = c Concave down means f’(c) is decreasing and f’’(c) < 0 ! f(x) is concave up at x = 1 but concave down at x = -1

Point of Inflection: a point where the graph of a function has a tangent line and changes concavity f must be continuous at a point of inflection So... To find local extrema (critical points), find when... f(x) changes slope (from + to – , or visa versa), or f’(x) changes sign (when f‘(x) = 0), or f‘(x) does not exist To find changes in concavity (pts. of inflection), find when... f‘(x) changes slope (from + to – , or visa versa), or f’’(x) changes sign (when f‘’(x) = 0 or f‘’(x) does not exist) If f is differentiable at all points, then... At a maximum, the graph is always concave down At a minimum, the graph is always concave up

- - + + + Ex: Find the local extrema of . Find the critical points... f‘ = 0 when x = -2/3 and x = 1, and f’ exists everywhere, so those are the 2 critical points Don’t forget to plug -2/3 and 1 into f to find the y-values! There is a local maximum at (-2/3, 130/27) and a local minimum at (1, 5/2). To verify the local extrema, one can use a 2nd derivative sign chart. Since f’’(-2/3) < 0 and f’(-2/3) = 0, f has a local maximum at x = -2/3 Since f’’(1) > 0 and f’(1) = 0, f has a local minimum at x = 1 - + + - +

- - + Ex: Find the local extrema and concavity of f‘ = 0 when x = -2 and 2 Also, f’ does not exist for , so consider those as endpoints Local min @ (-2, -4) and Local max @ (2, 4) and - - +

- + Ex (cont’d): Find the local extrema and concavity of Find the 2nd derivative! I’ll save you some time: x = 0 is the only point of inflection because we get 2x(x2 – 12) = 0. f is concave up from and concave down from . - +