Continuity And Differentiability

x-x² = x(1-x) f(x)= begincases 2x-x², & -1≤ x<0
x², & 0≤ x\le1 endcases Continuity: limₓtₒ₀⁻f(x)=limₓtₒ₀⁺f(x)=0=f(0) So f is continuous on [-1,1].
Differentiability at 0: f'_-(0)=2, f'_+(0)=0 Since derivatives are unequal, f is not differentiable at x=0.

Step 1: We are given the function .
To determine the intervals where this function is increasing, we need to find the derivative of .
The derivative of can be found using the chain rule:
Step 2: The derivative of is:
Therefore, the derivative of is:
Step 3: For to be increasing, .
This means that the numerator must be positive.
So, we need to solve:
Rewriting this inequality:

This inequality holds in the interval , because in this interval, is greater than .
Step 4: Thus, the function is increasing in the interval .

Step 1: Consider the given limit .
For (approaching 0 from the right), is positive, so .
The limit becomes: For (approaching 0 from the left), is negative, so . The limit becomes:
Since the limit from the right is 1 and the limit from the left is -1, the two one-sided limits are not equal.
Hence, the limit does not exist.

Step 1: To check if two lines are coplanar, we use the condition that the scalar triple product of the direction vectors of the lines and the vector joining a point on one line to a point on the other line must be zero.
The direction ratios of the first line are: , and for the second line, the direction ratios are: .
The point on the first line can be taken as , and the point on the second line can be taken as .
The vector joining these two points is:
We now calculate the scalar triple product:
For the lines to be coplanar, the scalar triple product must be zero: Thus, or .

Step 1: We are given the objective function and the constraints: To find the maximum value of , we will first graph the constraints and identify the feasible region, and then evaluate the objective function at the corner points (vertices) of the feasible region.
Step 2: Rewrite the constraints as equations: - (Line 1), - (Line 2). The feasible region is bounded by these lines and the axes.
Step 3: Find the intersection points of these lines:
- Intersection of and : Solve the system of equations: From Equation 1, .
Substitute into Equation 2: Substitute into :
Thus, the intersection point is .
- Intersection of and the x-axis (where ):
Thus, the point is .
- Intersection of and the y-axis (where ):
Thus, the point is .
Step 4: Now, evaluate at each corner point: - At , , - At , , - At , .
The maximum value of is , which occurs at .