Math

QuestionDetermine if the following functions are odd/even and injective/surjective/bijective:
1. f(x)=cf(x)=c
2. f(x)=xf(x)=x
3. f(x)=x2f(x)=x^{2}
4. f(x)=x3f(x)=x^{3}
5. f(x)=1xf(x)=\frac{1}{x}
6. f(x)=1x2f(x)=\frac{1}{x^{2}}
7. f(x)=x3f(x)=\sqrt[3]{x}
8. f(x)=xf(x)=\sqrt{x}
9. f(x)=xf(x)=|x|

Studdy Solution

STEP 1

Assumptions1. All functions are real-valued functions defined on the set of real numbers. . A function is even if f(x)=f(x)f(x) = f(-x) for all xx in the domain of ff.
3. A function is odd if f(x)=f(x)f(x) = -f(-x) for all xx in the domain of ff.
4. A function is injective (or one-to-one) if different inputs always give different outputs.
5. A function is surjective (or onto) if every possible output is produced by at least one input.
6. A function is bijective if it is both injective and surjective.

STEP 2

For f(x)=cf(x) = c, where cc is a constant, the function is even because f(x)=f(x)f(x) = f(-x) for all xx in the domain of ff. It is neither injective nor surjective.

STEP 3

For f(x)=xf(x) = x, the function is odd because f(x)=f(x)f(x) = -f(-x) for all xx in the domain of ff. It is both injective and surjective, hence bijective.

STEP 4

For f(x)=x2f(x) = x^2, the function is even because f(x)=f(x)f(x) = f(-x) for all xx in the domain of ff. It is not injective but it is surjective for x0x \geq0.

STEP 5

For f(x)=x3f(x) = x^3, the function is odd because f(x)=f(x)f(x) = -f(-x) for all xx in the domain of ff. It is both injective and surjective, hence bijective.

STEP 6

For f(x)=1/xf(x) =1/x, the function is odd because f(x)=f(x)f(x) = -f(-x) for all xx in the domain of ff. It is both injective and surjective, hence bijective.

STEP 7

For f(x)=1/x2f(x) =1/x^2, the function is even because f(x)=f(x)f(x) = f(-x) for all xx in the domain of ff. It is not injective but it is surjective for x0x \geq0.

STEP 8

For f(x)=f(x) = cube root of xx, the function is odd because f(x)=f(x)f(x) = -f(-x) for all xx in the domain of ff. It is both injective and surjective, hence bijective.

STEP 9

For f(x)=f(x) = square root of xx, the function is neither odd nor even. It is injective and surjective for xx \geq.

STEP 10

For f(x)=xf(x) = |x|, the function is even because f(x)=f(x)f(x) = f(-x) for all xx in the domain of ff. It is not injective but it is surjective for x0x \geq0.

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