Transcript AP Calculus Section 3.4 Concavity and the Second Derivative Test
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 decreasing on the interval.
I if f’ is Concave upward. f’ is increasing. The graph of f lies above its tangent lines.
Concave downward. f’ is decreasing. The graph of f lies below 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:
Theorem 3.7
Test for Concavity
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.
Can you explain why???
Example 1 (#3) Determine the open intervals on which 𝑓 𝑥 = downward.
24 𝑥 2 +12 is concave upward or concave Interval Test Value Sign of f’’(x) Conclusion
Example 2 (#10) Determine the open intervals on which y = 𝑥 + concave downward.
2 𝑠𝑖𝑛𝑥 on ( −𝜋, 𝜋) is concave upward or 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] .