AP Calculus AB: 8.4.4 The Second Derivative Test
This content explains how the second derivative test helps classify critical points as relative maxima or minima based on the concavity of the function at those points. It also covers cases where the test is inconclusive and the need for additional analysis using the first derivative.
The Second Derivative Test
• If the graph of a function has a tangent line with a slope of 0 and the graph is concave up at the same point, then the point is a minimum point of the function. If the graph of the function is concave down at that point, then the point is a maximum point of the function.
• If the graph of a function has a tangent line with a slope of 0 and the second derivative is at that point is also 0, then the second derivative test is inconclusive.
• The second derivative test states that if f (c) = 0 and the second derivative of f exists on an open interval containing c, then f(c) can be classified as follows:
1) If f (c) > 0, then f(c) is a relative minimum of f.
2) If f (c) < 0, then f(c) is a relative maximum of f.
3) If f (c) = 0, then the test is inconclusive.
Key Terms
The Second Derivative Test
• If the graph of a function has a tangent line with a slope of 0 and the graph is concave up at the same point, then the point is a minimum point ...
note
The second derivative test indicates whether a critical point is a maximum point or minimum point without making a sign chart for the first...
Given that the function f(x) has a critical point at x= e^1.7 and the second derivative is f′′(x)=0, what can be said about f(x) at x=e^1.7?
You need more information to determine if the point is a maximum or minimum
Given that the function f(x) has a critical point at x=π/2 and the second derivative is f′′(x)=−sin x, what can be said about f(x)at x=π/2?
The point is a maximum
Given that the function f (x) has a critical point at x = 3 and the second derivative is f ″(x) = 2, what can be said about f (x) at x = 3?
There is a minimum point there.
Given that the function f (x) has a critical point at x = −1 and the second derivative is f ″(−1) = −3, what can be said about f (x) at x = −1?
There is a maximum point there
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| Term | Definition |
|---|---|
The Second Derivative Test | • If the graph of a function has a tangent line with a slope of 0 and the graph is concave up at the same point, then the point is a minimum point of the function. If the graph of the function is concave down at that point, then the point is a maximum point of the function. |
note |
|
Given that the function f(x) has a critical point at x= e^1.7 and the second derivative is f′′(x)=0, what can be said about f(x) at x=e^1.7? | You need more information to determine if the point is a maximum or minimum |
Given that the function f(x) has a critical point at x=π/2 and the second derivative is f′′(x)=−sin x, what can be said about f(x)at x=π/2? | The point is a maximum |
Given that the function f (x) has a critical point at x = 3 and the second derivative is f ″(x) = 2, what can be said about f (x) at x = 3? | There is a minimum point there. |
Given that the function f (x) has a critical point at x = −1 and the second derivative is f ″(−1) = −3, what can be said about f (x) at x = −1? | There is a maximum point there |
If f(x) has a critical point at x=5,f′(5)=0, and f′′(x)=3x−5, what can be said about the function at x=5? | f(x) has a minimum point at x=5 |
Given that the function f(x)has a critical point at x=√2and that the second derivative is undefined at that point, what can be said about the function at x=√2? | No additional information can be determined. |