Real-Valued Functions, Their Domain and Range

Since \(x^2\ge0\), we have \(x^2+1\ge1\), so the denominator never becomes zero and the domain is \(\mathbb R\). Also, \(f(0)=1\) is the maximum value, while the function approaches 0 but never equals it. Exam tip: first check where the denominator is zero.

For x ≥ 1, the square root is non-negative, so y ≤ 7. At x = 1, y = 7. As x increases, √(x − 1) can grow without bound, so y has no lower bound. The range is therefore (−∞, 7].

For a real square root, x² − 16 must be non-negative. Thus x² ≥ 16, which gives x ≤ −4 or x ≥ 4. Therefore, the domain is (−∞, −4] ∪ [4, ∞).

First evaluate the expression inside the floor function: \(2.6+0.5=3.1\). The greatest-integer function \(\lfloor y\rfloor\) gives the greatest integer less than or equal to \(y\), so \(\lfloor3.1\rfloor=3\). Option B incorrectly takes the integer part of \(2.6\) without adding \(0.5\). Exam tip: evaluate the quantity inside the floor symbol before applying the function.

For x ∈ [1, 1.5), we have 2x ∈ [2, 3). The greatest integer less than or equal to any number in [2, 3) is 2, so f(x) = 2 throughout the domain. Thus, the range is {2}; 3 is not attained because x = 1.5 is excluded.

When \(x=7\), the input to the signum function is \(5-7=-2\). Since \(-2\) is negative, \(\operatorname{sgn}(-2)=-1\). The value 0 occurs only when the input is zero, while 1 is used for a positive input. Exam tip: first evaluate the expression inside the signum function and then identify its sign.

The composition means \((f\circ g)(x)=f(g(x))\). Hence, \((f\circ g)(x)=\sqrt{x-7}\). For a real square root, \(x-7\ge 0\), so \(x\ge 7\), giving the domain \([7,\infty)\). Option B is incorrect because it excludes 7, even though the function is defined at \(x=7\) with value 0. Exam tip: For the domain of a square-root expression, set the expression inside the root greater than or equal to zero.

The domain of a sum is the intersection of the domains of its functions. In f(x), the square root is in the denominator, so x − 4 must be strictly positive: x > 4. Since g is defined for all real numbers, the domain of (f + g)(x) is (4, ∞).

For \(\sqrt{x-1}\) to be defined, \(x-1\ge 0\), so \(x\ge 1\). In addition, the denominator \(g(x)=x-5\) must be nonzero, so \(x\ne 5\). Combining these conditions gives the domain \([1,\infty)\setminus\{5\}\). Option B is incorrect because \(x=1\) is allowed: the square root is \(0\) there. Exam tip: for a quotient, intersect the domains of the numerator and denominator functions, then exclude every value that makes the denominator zero.

A quotient function cannot be defined where its denominator is zero. Thus \(g(a)=0\) makes \(h(a)\) undefined, even if \(f(a)=0\). Exam tip: find denominator zeros first and exclude them from the domain.