Real-Valued Functions, Their Domain and Range

Correct answer: A. \(\{-1,0,1\}\)

For the signum function, \(\operatorname{sgn}(x)=-1\) for \(x<0\), \(\operatorname{sgn}(x)=0\) for \(x=0\), and \(\operatorname{sgn}(x)=1\) for \(x>0\). Hence, the function takes only the three values \(-1,0,1\), so its range is \(\{-1,0,1\}\). Option D omits \(-1\), which occurs for negative values of \(x\). Exam tip: evaluate the three cases \(x<0\), \(x=0\), and \(x>0\) for every signum-function question.

Correct answer: A. \(x\ge -2\)

For a real-valued square root, the radicand must be non-negative. Thus, \(x+2\ge 0\), which gives \(x\ge -2\). The domain is therefore \([-2,\infty)\).

Correct answer: C. 4

A square root is always non-negative, so \(\sqrt{x}\ge 0\). Therefore, \(f(x)=\sqrt{x}+4\ge 4\). At \(x=0\), \(f(0)=4\), so the minimum value is 4. The value 0 is the minimum of \(\sqrt{x}\) alone, not of the whole function. Exam tip: For a square-root function, first use the fact that the square-root term is at least zero.

Correct answer: C. \(x=2\)

The denominator of a fraction cannot be zero because division by zero is undefined. Here the denominator is \(x-2\). Setting \(x-2=0\) gives \(x=2\), so the function is not defined at \(x=2\). Exam tip: For a rational function, set its denominator equal to zero to find the excluded value from the domain.

Correct answer: A. \(x\ge 5\)

For a real square root, the expression under the radical must be non-negative, so \(x-5\ge 0\). Solving this inequality gives \(x\ge 5\), hence the domain is \([5,\infty)\). Option B is incorrect because values with \(x<5\) make the radicand negative. Exam tip: for a square-root function, set the radicand greater than or equal to zero.

Correct answer: A. \(x=3\)

The denominator of a fraction cannot be zero. Here, the denominator is \(x-3\). Setting \(x-3=0\) gives \(x=3\), so the function is undefined at \(x=3\). The other given values do not make the denominator zero. Exam tip: To find the domain of a rational function, set its denominator equal to zero and exclude the resulting value.

Correct answer: A. 1

Substituting \(x=5\) gives \(x-2=5-2=3>0\). The signum function has value \(1\) for a positive number, so \(f(5)=1\). The value \(0\) would occur only at \(x=2\), while \(-1\) occurs when \(x<2\). Exam tip: first determine the sign of the expression inside the signum function.

Correct answer: B. 0

When \(x=-1\), we get \(x+1=0\). For the signum function, \(\operatorname{sgn}(0)=0\); hence, \(f(-1)=0\). Exam tip: \(\operatorname{sgn}(t)\) equals 1 for \(t>0\), 0 for \(t=0\), and -1 for \(t<0\). Therefore, 1 and -1 are incorrect here, and the function is defined at \(x=-1\).

Correct answer: A. \(f(x)=e^x\)

\(e^x\) is defined for every real \(x\) and is always positive. For each \(y>0\), taking \(x=\ln y\) gives \(e^x=y\). \(x^2\) also includes 0 in its range. Exam tip: check domain restrictions in roots and fractions.

Correct answer: B. \(x=3\)

The zero of a function is the value of \(x\) for which the function equals zero. Thus, set \(3x-9=0\), giving \(3x=9\) and hence \(x=3\). The distractor \(x=9\) results from forgetting to divide by 3. Exam tip: To find the zero of a linear function, set \(f(x)=0\).

Correct answer: A. All real numbers

For the cubic function \(f(x)=x^3\), the cube of every real number is defined; there is no denominator, square root, or logarithm imposing a restriction. Hence, its domain is all real numbers. Options B and C unnecessarily exclude negative numbers or zero. Exam tip: the domain of every polynomial function is all real numbers.

Correct answer: A. \([-4,4]\)

For a square-root function to have a real value, its radicand must be non-negative. Therefore require \(16-x^2\ge0\). Rearranging gives \(x^2\le16\), which is equivalent to \(|x|\le4\), or \(-4\le x\le4\). The endpoints are included because at \(x=4\) and \(x=-4\) the radicand is zero and the square root is defined, giving output 0. Values outside this interval make \(16-x^2<0\), so they do not produce real outputs. Hence the real domain is \([-4,4]\), option A. Option D incorrectly allows all real numbers, while options B and C retain only one side of the required symmetric interval.

Correct answer: A. R\{-2}

The governing concept for a quotient function is that the denominator must be nonzero, while the numerator and denominator must otherwise be defined. Both given expressions are polynomials, so they are defined for all real x. The denominator is g(x)=x+2, and setting it equal to zero gives x=-2. This one value must be excluded, so the domain is R\{-2}. It is not necessary to exclude a zero of the numerator; a quotient can equal zero when its numerator is zero, provided the denominator is nonzero. Option B incorrectly includes the forbidden denominator value, option C removes the wrong number, and option D keeps only part of the valid domain. Therefore option A is correct.

Correct answer: C. 10

In f(x) = 1/(x - 10), the denominator cannot be zero because division by zero is undefined. From x - 10 = 0, we get x = 10. Therefore, x = 10 is excluded from the domain. Exam tip: for a rational function, set the denominator equal to zero to find excluded values.