Calculus

Step 1: Find the first derivative of .
The function is given by . The derivative is:
Step 2: Use the fact that extreme points occur where the first derivative is zero.
Given that and are extreme points, we have and . For : For :
Step 3: Solve the system of linear equations for and .
We have the system:
1)
2) (multiplying the second equation by 2)
Adding the two equations: Substitute into the first equation: Thus, and .

Step 1: Analyze the given conditions.
Since is a polynomial of least degree with a local maximum and minimum at and , respectively, the derivative must have roots at and . Thus, where is a constant.
Step 2: Find by integrating .
Integrate : where is the constant of integration.
Step 3: Use the given conditions.
Given: Substitute into : Substitute into :
Step 4: Solve for .
Substituting in equation (1):
Step 5: Find .
Now, Substituting : Thus, .

Step 1: Define and compute its derivative.
We are given , and we need to determine the behavior of . To check if is increasing or decreasing, compute its derivative: Using the chain rule for , let , so :
Step 2: Analyze on .
For , the argument . We know for , so . However, , and we have no direct information about on . To proceed, consider the symmetry of : This shows is symmetric about , but we focus on .
Step 3: Determine the sign of on .
Since , and , we need to compare and . Since on , is increasing on . For , assume may decrease or change behavior beyond . If is large near and is smaller near , then , making .
Step 4: Hypothesize and test.
Assume is linear, say , with since . Then: This implies is constant, not decreasing, so a linear doesn’t work. Try a concave function, say on , with for simplicity: On , , and at , so is decreasing on . This matches option (B).
Step 5: Generalize the behavior.
Since , is increasing on . If is concave ( ), then is decreasing. For , , so , making , confirming is decreasing on .

Step 1: Find the critical points.
To find local maxima and minima, compute the first derivative of : Set to find critical points: Solve the quadratic equation: so or . The critical points are and .
Step 2: Determine the nature of critical points using the second derivative.
Compute the second derivative: Evaluate at the critical points: - At : indicating a local maximum (since ). - At : indicating a local minimum (since ).
Step 3: Analyze the number of extrema.
The function is a cubic polynomial, and its derivative is a quadratic with two roots ( ). Thus, there are exactly two critical points:
: local maximum.
: local minimum.
This gives one local maximum and one local minimum.

Step 4: Verify against options.
% Option (A) Two local maxima: Incorrect, as there is only one local maximum.
% Option (B) Two local minima: Incorrect, as there is only one local minimum.
% Option (C) One local maximum and one local minimum: Correct, as found.
% Option (D) Neither maximum nor minimum: Incorrect, as extrema exist.