Functions as a special kind of relation

For the formula \\(f(x)=\\frac{1}{x^2+ax+4}\\) to define a real-valued function for every real x, its denominator must never be zero. Consider the quadratic denominator \\(x^2+ax+4\\). Since its leading coefficient is positive, it remains strictly positive for all real x only when it has no real zero. The discriminant must therefore be negative: \\(a^2-16<0\\).

Solving this inequality gives \\(a^2<16\\), which is equivalent to \\(-4<a<4\\), or \\(|a|<4\\). At \\(|a|=4\\), the denominator has one real zero, so the reciprocal is undefined at that point. If \\(|a|>4\\), it has two real zeros. Hence option A is correct. The condition does not hold for every real a; it must satisfy the strict inequality.

Choose exactly 3 of the 6 inputs to map to 1; the remaining 3 inputs must map to 0. Hence, the number of functions is \(\binom{6}{3}=20\). The value 64 counts all functions from a 6-element set to a 2-element set, without fixing how many inputs map to 1. Exam tip: for a function into \(\{0,1\}\) with exactly \(r\) inputs mapping to 1, the count is \(\binom{n}{r}\).

The rule sends (x) to (6-x), and applying the same rule again returns the original value. Thus the inverse relation is the same function.

Since (x^2-4x+7=(x-2)^2+3), the minimum value is (3). Complete the square to find the range of a quadratic function.

At (n=1), (\frac{n}{2}=\frac{1}{2}) is not a natural number. In (\mathbb{N}\to\mathbb{N}), every natural input must have a natural value.

On the given domain (f(x)=x+4), but removing (x=4) removes the value (8). After simplification, the image of the excluded input is removed.

With exactly (4) pairs, a function needs one image for each element of (A). Balance among second components is not required.

By symmetry the maximum occurs at (x=\frac{9}{2}), and the value is (2\sqrt{\frac{9}{2}}=3\sqrt{2}). For sums of square roots, check the balanced point.

The direct answer is A: there are \(2^{15}\) relations and \(5^3\) functions. A relation from \(A\) to \(B\) is any subset of the Cartesian product \(A\times B\). Since \(|A|=3\) and \(|B|=5\), the product has \(3\times5=15\) ordered pairs. Each pair may independently be included or excluded, giving \(2^{15}\) relations. A function is more restricted: each of the 3 domain elements must choose exactly one of the 5 codomain elements. Thus the number of functions is \(5\times5\times5=5^3\). Option A is correct. Option B gives 15, which is only the number of possible ordered pairs, not all subsets. Option C reverses the two counts. Option D uses incorrect sizes and formulas. Notice that every function is a relation, but most relations are not functions because they may omit an input or give one input several outputs. Memory cue: relations use \(2^{|A||B|}\); functions use \(|B|^{|A|}\).

At the given inputs, the values are \(-\frac{1}{2},-2,4,\frac{5}{2}\). For a finite domain, substitute directly.

The direct answer is A: \(f(2)=3\). In a piecewise function, first decide which condition contains the input, and only then use that formula. The first rule \(2x+1\) applies when \(x<2\), so it does not apply at \(x=2\). The second rule \(x^2-1\) applies when \(x\ge2\), and 2 satisfies this condition. Therefore \(f(2)=2^2-1=4-1=3\). Option A is correct. Option B, 4, is the intermediate value \(2^2\), before subtracting 1. Option C, 5, could come from using the wrong expression or an arithmetic error, but it is not the value of either applicable calculation. Option D is wrong because the second branch explicitly defines the function at 2. The boundary sign matters: \(x<2\) excludes 2, while \(x\ge2\) includes 2. Memory cue: at a boundary, read the inequality carefully before substituting.

The expression (\sqrt{2x-1}) is real only when (x\ge\frac{1}{2}). A function on all of (\mathbb{R}) needs a value for every real input.

There are (2^5=32) total functions, and the opposite case (f(1)=b), (f(5)=a) gives (2^3=8) functions. Hence (32-8=24).

The direct answer is A: the range is \([6,\infty)\). The expression \(|x+2|\) is the distance from x to -2, and \(|x-4|\) is the distance from x to 4. The distance between -2 and 4 is 6. If x lies between these points, the two distances add to exactly 6: for \(-2\le x\le4\), \(|x+2|+|x-4|=(x+2)+(4-x)=6\). Outside this interval, moving farther away increases the sum, so values larger than 6 occur. Hence the minimum is 6 and every value from 6 upward is possible; the range is \([6,\infty)\). Option A is correct. Option B incorrectly allows values below the minimum, including 0. Option C gives only the minimum, not all possible values. Option D includes values at most 6 and even negative values, impossible because absolute values are nonnegative. Memory cue: a sum of distances is at least the distance between the fixed points.

Each given (x) has a unique real fifth root in (Y). The inverse of an odd power is unique over real values.

Putting \(x=0\) into \(|x+y|=2\) gives \(|y|=2\), so both \(y=2\) and \(y=-2\) form ordered pairs in the relation. A function must assign exactly one image to each domain element; two images for the same \(x\) violate this condition. In fact, \(x=2\) gives \(y=0,-4\), and \(x=-2\) gives \(y=0,4\), so options B and C are also incorrect. Exam tip: to test a relation for being a function, check that every domain element has exactly one image.

There are (5), then (4), then (3), then (2) choices for the four inputs. Total functions are (5\cdot4\cdot3\cdot2=120).

In option (B), every input has one image, so (R) is a function. In the inverse relation, (4) has two images (1) and (2).

The direct answer is option A: 1. A function assigns exactly one output to each permitted input. Here the input is x=1. The rule is written in two parts: use x^2 when x≤1, and use x+2 only when x>1. Since 1≤1 is true, the first rule applies. Therefore f(1)=1^2=1. The second rule is not used, because 1 is not greater than 1. Option A is correct because it gives this value. Option B, 3, would come from incorrectly using 1+2, but that rule is reserved for inputs strictly greater than 1. Option C is wrong because a function cannot give two outputs to the same input; the boundary condition selects only the first rule. Option D is wrong because 1 is included in the first case, so the function is defined there. A useful exam cue is to check the equality sign carefully: ≤ includes the boundary, whereas > does not.

The domain includes (1), but (\frac{2}{1-1}) is undefined. A function needs a value for every domain element.

To find the range, first study the expression inside the square root. Completing the square gives x²+4x+8=(x+2)²+4. Since a square is always nonnegative, the smallest possible value of the expression is 4, reached at x=−2. As x varies over all real numbers, the expression can take every value from 4 upward. The square root therefore can take every value from √4=2 upward.

Hence the range is [2,∞), so option A is correct. The output cannot be below 2 because the quantity under the root is never below 4. Also, every number y≥2 is possible: choose x so that (x+2)²=y²−4. Thus there is no upper bound. The supplied answer correctly identifies both endpoints and the unbounded direction.

An absolute value expression assigns one non-negative value to each input. Equal images for different inputs do not stop it from being a function.

The values are (f(1)=5), (f(2)=2), (f(3)=3), (f(4)=2), and (f(5)=5). Therefore the range is ({2,3,5}).

The graph of a function has exactly one ordered pair for each element of the domain. Therefore the number of pairs is (|A|=6).

The expression (\sqrt{x-1}) gives the principal non-negative square root, so each (x) has one image. The obtained values are ({0,1,2,3}).