Real-Valued Functions, Their Domain and Range

The direct answer is A: \\( (-3,-2]\cup[2,3) \\). A square root requires the fraction to be non-negative, and the denominator cannot be zero. Thus \\( (x^2-4)/(9-x^2)\ge0\\), with x not equal to ±3. Factor: \\(x^2-4=(x-2)(x+2)\\) and \\(9-x^2=(3-x)(3+x)\\). The critical points are -3,-2,2,3. Testing intervals gives positive values on \\((-3,-2]\\) and \\([2,3)\\); the zeros -2 and 2 are included, while -3 and 3 are excluded because the denominator is zero. A is correct. B includes values between -2 and 2, where the fraction is negative. C uses outside intervals, where the fraction is negative. D includes forbidden points and also intervals with negative radicand. Exam cue: include numerator zeros, exclude denominator zeros, then use a sign chart.

Let (t=x^2\ge 0), then (f=\frac{2t+3}{t+1}) is decreasing. At (t=0), (3) is obtained and (2) is not attained.

For the square root to be real, (x − 2)(6 − x) must be at least zero. The factors show that this product is non-negative from x = 2 through x = 6, so the radical condition gives [2, 6]. The denominator adds the restriction x − 4 ≠ 0, meaning x ≠ 4. Removing 4 from [2, 6] gives [2, 4) ∪ (4, 6]. The endpoints 2 and 6 remain included because the radicand is zero there and the denominator is nonzero. Therefore option A is correct. Option B forgets the denominator restriction, C wrongly excludes the valid endpoints, and D includes many real numbers for which the radicand is negative.

Because log₅x is under a square root in the denominator, two conditions are required: the logarithm must be defined and the square-root denominator must be strictly positive. The logarithm first requires x > 0. More importantly, √(log₅x) cannot be zero, so log₅x > 0. Since the base 5 is greater than 1, the logarithm is positive exactly when x > 1. Therefore the domain is (1, ∞), which is option C. Option A includes 0 < x ≤ 1, where log₅x is non-positive; option B incorrectly includes x = 1, making the denominator zero; and option D contains values for which log₅x is negative, so the square root is not real.

Rewrite the function as f(x) = x + 4/x, with x ≠ 0. To find its range, set y = x + 4/x and multiply by x: x² − yx + 4 = 0. A real x exists exactly when the discriminant is nonnegative: y² − 16 ≥ 0. Thus y ≤ −4 or y ≥ 4. The endpoint values are attainable: f(2) = 4 and f(−2) = −4. Therefore the range is (-∞, −4] ∪ [4, ∞), so option A is correct. Option B includes only the positive branch, C incorrectly assumes every nonzero real value occurs, and D lists values that cannot occur. The same result also follows from AM-GM separately on positive and negative x.

First rewrite the function as f(x)=x+1/x, with x≠0. To determine its range, let y=x+1/x. For x>0, the AM-GM inequality gives x+1/x≥2, with equality at x=1. Thus the positive branch has range [2,∞). For x<0, applying the corresponding inequality to −x shows x+1/x≤−2, with equality at x=−1. Thus the negative branch has range (-∞,−2]. Combining both branches gives (-∞,−2]∪[2,∞), option A. Values strictly between −2 and 2 cannot occur. Option B omits the negative branch, option C incorrectly excludes only zero, and option D lists precisely values that are not in the range.

When \(x=0\), \(|x|=0\), so the denominator of \(\frac{x}{|x|}\) is zero. Hence, the function is undefined at this point. For \(x=1,-1\), and \(2\), the denominator is non-zero, so the function is defined. Exam tip: While finding the domain of a rational expression, exclude every value that makes its denominator zero.

For \(f(x)=x^2+1\), every real \(x\) is allowed, so the domain is \(\mathbb{R}\). Since \(x^2\ge0\), \(f(x)\ge1\), giving range \([1,\infty)\). Unlike this, \(x^3\) has range \(\mathbb{R}\). Exam tip: first locate the minimum of a squared term.

The product is \\((fg)(x)=\\frac{\\sqrt{x}}{\\sqrt{4-x}}\\). The function \\(f(x)\\) requires \\(x\\ge0\\). Since \\(\\sqrt{4-x}\\) is in the denominator, it must be strictly positive, so \\(4-x>0\\), or \\(x<4\\). The intersection of these conditions is \\(0\\le x<4\\), making option A correct. Option C misses the necessary restriction \\(x<4\\). Exam tip: For a product or quotient of functions, find the domain by intersecting all individual domain restrictions and excluding zero denominators.

\((f-g)(x)=\sqrt{x-1}-\sqrt{3-x}\) is defined only when both square roots are real. Thus, \(x-1\ge0\) gives \(x\ge1\), and \(3-x\ge0\) gives \(x\le3\). Their intersection is \(1\le x\le3\), so option A is correct. Options B and C use the condition from only one square root and are therefore incomplete. Exam tip: for the sum or difference of functions, the domain is the intersection of the domains of all component functions.

Since \(x^2-1=(x-1)(x+1)\), for \(x\ne -1\) we get \(f(x)=\frac{(x-1)(x+1)}{x+1}=x-1=g(x)\). However, the original function \(f\) is undefined at \(x=-1\) because its denominator becomes zero. Therefore, the functions have equal values for every real \(x\ne -1\). Exam tip: after simplifying a rational expression, always retain the domain restriction from the original expression.

\(x^3-x\) is a cubic polynomial, so both its domain and range are \(\mathbb{R}\). It is not one-one because \(f(-1)=f(0)=0\). Exam tip: find two distinct inputs with the same output.

For the product (fg)(x)=f(x)g(x), both factors must be defined. The logarithm log(x−2) requires a strictly positive argument, so x−2>0 and hence x>2. The square root √(8−x) requires a non-negative radicand, so 8−x≥0 and hence x≤8. The domain is the intersection of these conditions: x>2 together with x≤8, which is (2,8]. The endpoint 2 is excluded because log 0 is undefined, whereas 8 is included because √0 is defined. Thus option A is correct. Option B incorrectly includes 2, option D ignores the square-root restriction, and option C reverses both relevant restrictions.