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

2. 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

3. 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

4. - - + + + 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 - + + - +

5. - - + 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 (-2, -4) and
Local (2, 4) and - - +

6. - + 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 - +