What does the Second Derivative Test theorem state?
If $f'(c) = 0$ and $f''(c) > 0$, then $f(c)$ is a local minimum. If $f'(c) = 0$ and $f''(c) < 0$, then $f(c)$ is a local maximum.
How does the Second Derivative Test relate to concavity?
The sign of the second derivative determines the concavity of the function at a point, which helps determine if it's a local max or min.
What is the Intermediate Value Theorem?
If a continuous function, f, attains two values, it must also attain all values in between.
What is the Extreme Value Theorem?
If a function is continuous on a closed interval, it must have a maximum and minimum on that interval.
What is Rolle's Theorem?
If a differentiable function has equal values at two points, there must be a point between them where the derivative is zero.
What is the Mean Value Theorem?
There exists a point where the instantaneous rate of change (derivative) equals the average rate of change over an interval.
What is the Fundamental Theorem of Calculus?
It connects differentiation and integration, stating that they are inverse operations.
How does the second derivative test relate to the first derivative?
The second derivative test uses the first derivative to find critical points and then tests the second derivative at those points.
What is the relationship between the second derivative and inflection points?
Inflection points occur where the second derivative changes sign, indicating a change in concavity.
How can you use the second derivative test to find global extrema?
The second derivative test only finds local extrema. To find global extrema, you must also check endpoints and any points where the derivative is undefined.
Define a critical point.
A point where f'(x) = 0 or f'(x) does not exist.
What is a local maximum?
A point where the function's value is greater than or equal to the values at all nearby points.
What is a local minimum?
A point where the function's value is less than or equal to the values at all nearby points.
Define concavity.
The direction in which a curve bends. Concave up means the curve opens upwards; concave down means it opens downwards.
What is an inflection point?
A point on a curve where the concavity changes (from up to down or vice versa).
What does the second derivative tell us?
The concavity of the function and helps determine local extrema.
Define the Second Derivative Test.
A method using the second derivative to determine whether a critical point is a local maximum or minimum.
What does f''(x) > 0 imply?
The function is concave up.
What does f''(x) < 0 imply?
The function is concave down.
What does 'inconclusive' mean in the context of the Second Derivative Test?
The test fails to determine whether the critical point is a local max, min, or neither. Further analysis is needed.
How to find local extrema using the Second Derivative Test?
1. Find critical points. 2. Compute the second derivative. 3. Evaluate the second derivative at each critical point. 4. Determine if each point is a local max, min, or neither.
Steps to determine concavity of a function.
1. Find the second derivative, $f''(x)$. 2. Set $f''(x) = 0$ and solve for x. 3. Create a sign chart for $f''(x)$. 4. Determine intervals of concave up ($f''(x) > 0$) and concave down ($f''(x) < 0$).
How to find inflection points?
1. Find the second derivative, $f''(x)$. 2. Set $f''(x) = 0$ and solve for x. 3. Check if the concavity changes at each potential inflection point.
What to do if the Second Derivative Test is inconclusive?
Use the First Derivative Test or analyze the behavior of the function around the critical point.
How to use the second derivative to determine if $x = c$ is a local max?
Find $f'(x)$, set $f'(c) = 0$ to confirm c is a critical point. Then, find $f''(x)$ and evaluate $f''(c)$. If $f''(c) < 0$, then $x = c$ is a local max.
How to use the second derivative to determine if $x = c$ is a local min?
Find $f'(x)$, set $f'(c) = 0$ to confirm c is a critical point. Then, find $f''(x)$ and evaluate $f''(c)$. If $f''(c) > 0$, then $x = c$ is a local min.
How to find the global extremum given one critical point?
If the function is continuous and has only one critical point, and that point is a local extremum, then it's also the global extremum.
How to find critical points for $f(x) = 4sin(x)$ on $0 < x < 2\pi$?
1. Find $f'(x) = 4cos(x)$. 2. Set $4cos(x) = 0$. 3. Solve for x, which gives $x = \frac{\pi}{2}, \frac{3\pi}{2}$.
Given $f(x) = 4sin(x)$ and critical point $x = \frac{\pi}{2}$, how to determine if it's a local max or min?
1. Find $f''(x) = -4sin(x)$. 2. Evaluate $f''(\frac{\pi}{2}) = -4$. 3. Since $f''(\frac{\pi}{2}) < 0$, it's a local max.
How to determine the nature of critical points when $f'(x) = \frac{247}{x}$?
1. Set $f'(x) = 0$ to find critical points, but there are none. 2. Check where $f'(x)$ is undefined, which is at $x=0$. 3. Since $f(x)$ is not defined at $x=0$, there are no local extrema.
