The square root of a negative value is imaginary, so the value of x – 5 cannot be negative. Since the square root is the denominator of a fraction, it cannot be equal to 0 either. The value of x – 5 must be greater than 0; \(\sqrt{x-5}\) > 0, \(\sqrt{x-5}\) > 0, x > 5. The domain of f(x) is all real numbers greater than 5. Since the denominator of the fraction must always be positive, and the numerator of the fraction is –1, the value of f(x) will always be negative. No value of x will make f(x) = 0, so the range of f(x) is all real numbers less than 0.

The square root of a negative value is imaginary, so the value of 4x – 1 must be greater than or equal to 0. 4x – 1 ≥ 0, 4x ≥ 1, x ≥ ¹⁄₄. The domain of f(x) is all real numbers greater than or equal to ¹⁄₄. Since x must be greater than or equal to ¹⁄₄, the smallest value of f(x) is the square root of 0, which is 0. The range of the function is all real numbers greater than or equal to 0.

The domain of the function is all real numbers; any real number can be substituted for x. There are no x values that can make the function undefined. The range of a function is the set of all possible outputs of the function. Since the x term is squared, then made negative, the largest value that this term can equal is 0 (when x = 0). Every other x value will result in a negative value for f(x). The range of f(x) is all real numbers less than or equal to 0.

The range of a function is the set of all possible outputs of the function. All real numbers can be substituted for x in the function f(x) = x² – 4, so the domain of the function is all real numbers. Since the x term is squared, the smallest value that this term can equal is 0 (when x = 0). Therefore, the smallest value that f(x) can have is when x = 0. When x = 0, f(x) = 02 – 4 = –4. The range of f(x) is all real numbers greater than or equal to –4.

The domain of a function is the set of all possible inputs to the function. All real numbers can be substituted for x in the function f(x) = \(\frac{1}{(x^{2}-9)}\) excluding those that make the fraction undefined. Set the denominator equal to 0 to find the x values that make the fraction undefined. These values are not in the domain of the function; x² – 9 = 0, (x – 3)(x + 3) = 0, x – 3 = 0, x = 3; x + 3 = 0, x = –3. The domain of f(x) is all real numbers excluding 3 and –3.

The definition of the function states that the term before the ^ symbol should be squared and the term after the ^ symbol should be subtracted from that square. Begin with the innermost function: y^y: Square the second term, y, and subtract the first term, y: y² – y. Now evaluate y^(y² – y). Again, square the second term, y² – y, and subtract the first term, y: (y² – y)² – y = y⁴ – y³ – y³ + y² – y = y⁴ – 2y³ + y² – y.

The definition of the function states that the term before the ? symbol should be subtracted from the term after the ? symbol, and that difference should be divided by the sum of the two terms. Begin with the innermost function: a?b, which is given as a \(\frac{b-a}{a+b}\). Substitute the values of a and b: a \(\frac{b-a}{a+b}=\frac{-5-6}{6+(-5)}=\frac{-11}{1}=-11\). Now evaluate a?–11. Remember, the term before the ? symbol should be subtracted from the term after the ? symbol, and that difference should be divided by the sum of the two terms. Substitute the value of a again: \(\frac{-11-6}{6+(-11)}=\frac{-17}{-5}=\frac{17}{5}\).

The definition of the function states that the term after the % symbol should be raised to the power of the term before the % symbol.Replace the expression j%k with its given value, kj, and the problem becomes k%(k^j). The term after the % symbol, k^j, should be raised to the power of the term before the % symbol, k: (k^j)k = k^jk. Remember, when an exponent is raised to an exponent, multiply the exponents.

The definition of the function states that the term before the & symbol should be divided by the term after the & symbol, and that quotient should be added to the product of the two terms. Therefore, the value of q&p = ^p⁄_q + qp. Substitute 4 for p and –2 for q in this definition:–²⁄₄ + (–2)(4) = –¹⁄₂ – 8 = –8.5.

The definition of the function states that you must multiply by 3 the term before the @ symbol, then subtract from that product the term after the @ symbol. Begin with the function in parentheses, w@z, which is given as 3w – z. Replace the expression w@z with 3w – z, and the problem becomes (3w – z)@z. Again, multiply the term before the @ symbol by 3, and subtract from it the term after the @ symbol. Multiply 3w – z by 3 and subtract z: (3w – z)@z = 3(3w – z) – z = 9w – 3z – z = 9w – 4z.