Functions

The given function is:

$ f(x) x x = 0 x = 0 f(0) = 0 $.

Step 1: Complete the square to rewrite the quadratic function. Complete the square:

Step 2: Analyze the function's behavior.

- The function is a parabola that opens upwards (since the coefficient of is positive). - The minimum value occurs at , since the vertex of the parabola is at . At : Step 3: Determine the range.

Since the parabola opens upwards and the minimum value at is 1, the range of is .

The range of is [1, ∞), so the correct option is (B).

We need to check the continuity and differentiability of at .
Step 1: Checking Continuity at }
For continuity, we need to check if the left-hand limit, right-hand limit, and the value of the function at are equal.
Left-hand limit as :
Right-hand limit as :
Value of the function at : Since both the left-hand and right-hand limits are equal and , the function is continuous at .
Step 2: Checking Differentiability at }
To check differentiability, we need to verify if the left-hand and right-hand derivatives at are equal.
Left-hand derivative (for ):
Right-hand derivative (for ): At , the right-hand derivative is: Since the left-hand derivative is and the right-hand derivative is , they are not equal.
Therefore, the function is not differentiable at .

Thus, the function is continuous but not differentiable at , so the correct answer is (B) continuous and non-differentiable.

To determine where the function is one-to-one, we need to examine its monotonicity.
A function is one-to-one if it is either strictly increasing or strictly decreasing. First, compute the derivative of : Using the chain rule: To determine the intervals where the function is one-to-one, we look for intervals where is either always positive or always negative.
This will tell us where the function is strictly increasing or decreasing. We notice that the expression has a periodicity of , meaning it repeats every .
By examining the values of over one period, we find that the function is one-to-one in the interval , as the derivative does not change sign within this interval.

Therefore, the function is one-to-one in the interval .

We are given the piecewise function defined as:

We are asked to find:

Step 1: Find
Since , we use the third case of the piecewise function, :

Step 2: Find
Since , we use the second case of the piecewise function, :

Step 3: Find
Since , we use the first case of the piecewise function, :

Step 4: Add the values

Therefore, the correct answer is:
So, the correct answer is (D): 11

We are given the following:

• Cost function:
• Revenue function:

To find the value of (the number of items) that will give a profit, we need to calculate the profit function. The profit function is the difference between revenue and cost:

Substituting the given functions:

Simplifying:

To earn a profit, the profit must be greater than 0:

Solving for :

Therefore, the value of to earn a profit is greater than 50.

The correct answer is (A):

We are asked to find the domain of the function .

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

For a rational function (a fraction), the function is undefined when the denominator is equal to zero.

In this case, the denominator is .

We set the denominator equal to zero and solve for x:

Add to both sides:

The equation means that x can be either 1 or -1.

These are the values of x for which the function is undefined.

Therefore, the domain of the function is the set of all real numbers except for -1 and 1.

In set notation, this is written as or .

In interval notation, the domain is .

Comparing this with the given options:

• R - [-1, 1] (Incorrect, excludes the interval between -1 and 1)
• (-∞, 1) (Incorrect, excludes -1 and values greater than 1)
• (-∞, 1) ∪ (0, 1) (Incorrect, simplifies to (-∞, 1))
• R - {-1, 1} (Correct)

The correct domain is R - {-1, 1}.

1. Understand the problem:

Given the functional equation for all , with , , and , we need to find .

2. Solve the functional equation:

The equation is satisfied by exponential functions of the form:

Given , which is consistent.

3. Determine the constant :

From :

From :

Thus, .

4. Find :

Using :

However, this contradicts the given . Let's re-evaluate:

The general solution is . Given , , and , we have:

Correct Answer: (C) 6

1. Understand the problem:

We are given the function and need to find the pre-images of 17 and -3, i.e., the set of all such that and .

2. Find the pre-image of 17:

Solve :

Thus, the pre-image is .

3. Find the pre-image of -3:

Solve :

This has no real solutions, so the pre-image is the empty set .

Correct Answer: (C)