Functions as a special kind of relation

The input (x=1) belongs to both parts and gives different values (4) and (2). Two different outputs for one input do not define a function.

At (x=3), (2\cdot3-1=5) and (3+2=5), so there is no conflict. Overlap is valid when both values agree.

There are (3\cdot3) choices for the two equal groups and (3\cdot2) choices for the unequal last pair. Total functions are (3\cdot3\cdot3\cdot2=54).

If (a<0), the radicand can become negative for large (|x|), and if (a\ge0), its minimum is (9-6a). Thus (9-6a\ge0) gives (0\le a\le\frac{3}{2}).

For an integer (n), (n^5-n) is always divisible by (5). Therefore every value lies in (\mathbb{Z}).

For every (x), (y=12-2x) is unique and lies in the given codomain. Listing all images is a safe method for finite sets.

On the given domain (f(x)=x+2), but removing (x=-2) removes the value (0). After simplification, note the image of the excluded input.

There are (10) choices for the triple ((f(1),f(2),f(3))) and (4) choices for (f(4)=f(5)). Total is (10\cdot4=40).

For (-2\le x\le4), the value is (6), and at endpoints (x=-3,5), it is (8). Check modulus breakpoints and interval endpoints.

There is one empty function from the empty domain, but its graph has no ordered pair. Keep the count and graph size separate.

All elements of (C) lie in (A), and their images lie in the codomain (B). Hence (f(C)\subseteq B) is always true.

The correct answer is option A: x=0 has two images. A relation is a function from X to Y only when every element of X has exactly one image in Y. Use x^2+y^2=9 and test the allowed values. For x=0, y^2=9, so y=3 or y=-3. Thus the same input 0 is related to two outputs, and the relation is not a function. For x=3, 9+y^2=9, so y=0 only; therefore option B is wrong. For x=-3, the same calculation gives y=0, so x=-3 does have an image; option C is wrong. Option D is wrong because the presence of 0 in Y does not by itself prevent a relation from being a function. What matters is the number of images for each input. Option A correctly identifies the failure of the one-input-one-output rule. Memory cue: in a function, each x gets one and only one y; different x-values may share a y, but one x cannot have two y-values.

The function is defined for all real \(x\) only if the denominator never becomes zero. So we need \(x^2+ax+9\ne 0\) for all \(x\). A quadratic \(x^2+ax+9\) never equals \(0\) for real \(x\) exactly when its discriminant is negative: \(a^2-4\cdot 1\cdot 9=a^2-36<0\). Thus \(a^2<36\Rightarrow -6<a<6\).

We must choose the 4 elements of the 7-element domain that will be mapped to 1. This can be done in \(\binom{7}{4}=35\) ways. The remaining 3 elements must be mapped to 0, so they introduce no additional choices. Hence the number of functions is 35. Exam tip: For a function from an \(n\)-element set to \(\{0,1\}\), if exactly \(r\) domain elements map to 1, the count is \(\binom{n}{r}\).

The rule sends (x) to (7-x), and applying the same rule again returns (x). Therefore the inverse relation is the same function.

On the given domain (f(x)=x+5), but removing (x=5) removes the value (10). The image of the excluded input is removed from the range.

With exactly (5) pairs, a function needs one image for every element of (A). Repetition of second components is allowed.

By symmetry the maximum occurs at (x=8), and the value is (2\sqrt{8}=4\sqrt{2}). For sums of square roots, check the balanced point.

Total relations are (2^{|A||B|}=2^{12}), and total functions are (|B|^{|A|}=3^4). The formulas for relations and functions are different.

At the given inputs, the values are \(\frac{7}{2},5,-1,\frac{1}{2}\). For a finite domain, direct substitution is fastest.

There are (2^6=64) total functions, and the opposite case (f(2)=b), (f(5)=a) gives (2^4=16) functions. Hence (64-16=48).

The direct answer is option A: [7,∞). The expression is f(x)=|x+5|+|x−2|. Absolute value means distance on the number line: |x+5| is the distance from x to −5, and |x−2| is the distance from x to 2. The distance between −5 and 2 is 7. For every x between −5 and 2, the two distances add to exactly 7. Outside that interval, moving farther away increases the total distance, so values larger than 7 occur. Thus the minimum is 7 and the range is [7,∞). Option A is correct because 7 is included and every larger value is possible. Option B, [0,∞), wrongly assumes the sum can be zero; both distances cannot be zero because x cannot equal −5 and 2 simultaneously. Option C, {7}, includes only the minimum and ignores larger values outside the interval. Option D contains values below 7, which are impossible. The exam idea is that the sum of distances from a point to two fixed points is at least the distance between those points.

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

The direct answer is B: no, this is not a valid function from the stated domain to the stated codomain. A function must assign every input exactly one output, and that output must belong to the codomain. Check the inputs one by one: f(0)=0^2−3(0)+2=2, f(1)=1−3+2=0, f(2)=4−6+2=0, f(3)=9−9+2=2, and f(4)=16−12+2=6. Although each input has one value, 6 is not in {0,1,2,3,4}; therefore the mapping fails as a function into the given codomain. Option A is wrong because not all values lie in the codomain. Option B is correct because f(4)=6 is outside it. Option C is wrong because f(1)=0 is allowed and does not cause a problem. Option D is wrong: a polynomial can be evaluated on a finite set, but its outputs must still lie in the declared codomain. Memory cue: check every input and then check whether every output belongs to the codomain.

The direct answer is A. A relation is a function only when every domain element has exactly one image. For x=2, substitute into |x−2y|=2: |2−2y|=2. This means 2−2y=2 or 2−2y=−2. The first equation gives y=0, and the second gives y=2. Thus the same input x=2 has two images, 0 and 2, so the relation is not a function. Option A is correct for precisely this reason. Option B is wrong because x=0 does have images: |0−2y|=2 gives y=1 or y=−1, both in the codomain. Option C is wrong because x=4 also has images: |4−2y|=2 gives y=1 or y=3. Option D is irrelevant and wrong; the mere presence of y=3 in the codomain does not violate the function rule. The problem is two outputs for one input, not the size of the codomain. Memory cue: a function allows many inputs to share one output, but never one input to have two outputs.