High School

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Match each function with the appropriate category. For each example, the form of the function is given, along with the relevant domain \(A\) and codomain \(B\).

1. \(f(x) = 3x^2 - 5\), \(A = [0, +\infty)\), \(B = [-10, +\infty)\)
- Choose...
- Bijective
- Neither injective nor surjective
- Surjective but not injective
- Injective but not surjective

2. \(h(x) = \sin(x)\), \(A = (-180^\circ, 180^\circ)\), \(B = [-3, 3]\)
- Choose...
- Bijective
- Neither injective nor surjective
- Surjective but not injective
- Injective but not surjective

3. I don't know

Answer :

f(x) = 3[tex]x^{2}[/tex] - 5, A = [0, +∞), B = [-10, +∞): Neither injective nor surjective. h(x) = sin(x), A = (-180°, 180°), B = [-1, 1]: Surjective but not injective.

For a function to be classified as injective, or one-to-one, it means that each element in the domain maps to a unique element in the codomain. In the case of f(x) = 3[tex]x^{2}[/tex] - 5, since the function is a quadratic function, it is not one-to-one.

Different values of x can result in the same output, violating the injectivity condition. Similarly, the function is not surjective, as it does not cover the entire codomain B. The range of f(x) is [−5, +∞), which is a subset of B.

On the other hand, the function h(x) = sin(x) is surjective but not injective. Since the sine function has a periodic nature, multiple values of x can produce the same output in the range [-1, 1], making it not injective. However, it covers the entire range [-1, 1] and therefore is surjective.

In summary, f(x) = 3[tex]x^{2}[/tex] - 5 is neither injective nor surjective, while h(x) = sin(x) is surjective but not injective.

Learn more about injective here:

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