AP Calculus Section 3.4 Concavity and the Second Derivative Test Student Notes Definitions, Theorems, and Examples

Definition 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. f’ is decreasing. The graph of f lies below its tangent lines. Concave upward. f’ is increasing. The graph of f lies above its tangent lines.

The graph of a function f is concave upward at the point (c, f(c)) if f(c) exists and if for all x in some open interval containing c, the point (x, f(x)) on the graph of f lies above the corresponding point on the graph of the tangent line to f at c. Imagine holding a ruler along the tangent line through the point (c, f(c)): if the ruler supports the graph of f near (c, f(c)), then the graph of the function is concave upward. The graph of a function f is concave downward at the point (c,f(c)) if f(c) exists and if for all x in some open interval containing c, the point (x, f(x)) on the graph of f lies below the corresponding point on the graph of the tangent line to f at c. In this situation, the graph of f supports the ruler.

You try:

Test for Concavity Theorem 3.7 Let f be a function whose second derivative exists on an open interval I. 1.) If f’’(x) > 0 for all x 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. To apply Theorem 3.7, locate the x-values at which f’’(x) = 0 or f’’ does not exist. Then, use these x-values to determine test intervals. Finally, test the sign of f’’(x) in each of the test intervals.

A straight line is neither concave upward nor concave downward A straight line is neither concave upward nor concave downward. Can you explain why???

Example 1 (#3) Determine the open intervals on which 𝑓 𝑥 = 24 𝑥 2 +12 is concave upward or concave downward. Interval   Test Value Sign of f’’(x) Conclusion

Example 2 (#10) Determine the open intervals on which y =𝑥+ 2 𝑠𝑖𝑛𝑥 on (−𝜋, 𝜋) is concave upward or concave downward. Interval   Test Value Sign of f’’(x) Conclusion

Definition Let f be a function that is continuous on an open interval and let c be a point in the interval. If the graph of f has a tangent line at this point (c, f(c)), then this point is a point of inflection of the graph of f if the concavity of f changes from upward to downward (or downward to upward) at the point.

The graph of a function crosses its tangent line at a point of inflection.

Example 3 A (#16) Determine the points of inflection and discuss the concavity of the graph of 𝑓 𝑥 = 𝑥 3 (𝑥−4). Interval   Test Value Sign of f’’(x) Conclusion

Example 3 B (#24) Determine the points of inflection and discuss the concavity of the graph of 𝑓 𝑥 =𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥 on [0, 2𝜋]. Interval   Test Value Sign of f’’(x) Conclusion

Theorem 3.9 Second Derivative Test 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 has a relative minimum at (c, f(c)). 2.) If f’’(c) < 0, then f has a relative maximum at (c, f(c)). If f’’(c) = 0, the test fails. f may have a relative maximum, a relative minimum, or neither. In these cases, you can use the First Derivative Test.

Example 4 A (#34) Use the Second Derivative Test to find all relative extrema of 𝑔 𝑥 =− 1 8 (𝑥+2 ) 2 (𝑥−4 ) 2 .

Example 4 b (#40) Use the Second Derivative Test to find all relative extrema of 𝑓 𝑥 =2𝑠𝑖𝑛𝑥+𝑐𝑜𝑠2𝑥 on [0, 2𝜋].

Assignment – Section 3.4 p. 195-197 #5, 9, 15, 19, 26, 35, 37, 67 or 68 (you choose)