All AP Calculus BC Resources
Example Questions
Example Question #301 : Ap Calculus Bc
Example Question #302 : Ap Calculus Bc
Example Question #303 : Ap Calculus Bc
Example Question #31 : Derivatives
Example Question #32 : Derivatives
Example Question #33 : Derivatives
Example Question #34 : Derivatives
Example Question #35 : Derivatives
Find the points of inflection on the function's domain:
None of the other answers
There are no points of inflection
The points of inflection are the points at which a function's second derivative changes in sign.
To start, we must find the function's second derivative:
The derivatives were found using the following rules:
,
Next, we find the values at which the second derivative is equal to zero:
Using this value as a bound, we create intervals on which to evaluate the sign of the second derivative:
.
Note that at the bounds of the intervals, the second derivative is neither positive nor negative.
Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is negative, while on the second interval, the second derivative is positive. Thus, a point of inflection exists at , because the second derivative did change sign at this point.
Example Question #37 : Derivatives
At what value of does the function
have an inflection point?
To find inflection points (if there are any), we compute the second derivative of and attempt to set it to . In this case:
so
.
Note: To confirm this is an inflection point (where the concavity of the function actually changes sign ––not simply touches without changing sign––) we can take the derivative yet again:
,
so of course, at our point of interest where ,
.
Since is nonzero and negative, we know that the second derivative changes sign from positive to negative, and so at the function changes from concave up, to concave down.
Example Question #101 : Derivative At A Point
Calculate the derivative of at the point .
There are 2 steps to solving this problem.
First, take the derivative of .
Then, replace the value of x with the given point.
For example, if , then we are looking for the value of , or the derivative of at .
Calculate
Derivative rules that will be needed here:
- Derivative of a constant is 0. For example,
- Taking a derivative on a term, or using the power rule, can be done by doing the following:
Then, plug in the value of x and evaluate
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