1. **Problem statement:** Given sets $A = B = \{x \mid -2 \leq x \leq 2\}$ and functions: a) $f(x) = |x|$ b) $g(x) = \cos\left(\frac{\pi x}{2}\right)$ c) $h(x) = x^2 h$ Determine whether each function is injective, surjective, or bijective. 2. **Recall definitions:** - Injective (one-to-one): No two distinct inputs map to the same output. - Surjective (onto): Every element in $B$ has a pre-image in $A$. - Bijective: Both injective and surjective. ----- ### a) $f(x) = |x|$ 3. **Injectivity:** - Consider $x = 1$ and $x = -1$ in $A$. - $f(1) = 1$ and $f(-1) = 1$, so distinct inputs map to the same output. - Thus, $f$ is **not injective**. 4. **Surjectivity:** - The codomain $B = [-2, 2]$. - The range of $f$ is $[0, 2]$ because absolute value is always non-negative. - Negative numbers in $B$ (like $-1$) are not in the range, so $f$ is **not surjective**. 5. **Conclusion:** $f$ is **neither injective nor surjective**, so **not bijective**. ----- ### b) $g(x) = \cos\left(\frac{\pi x}{2}\right)$ 6. **Injectivity:** - The function $g$ is the cosine function scaled and shifted. - Since cosine is periodic and even, distinct inputs can share the same output. - For example, $g(-2) = \cos(-\pi) = -1$ and $g(2) = \cos(\pi) = -1$. - Hence, $g$ is **not injective**. 7. **Surjectivity:** - Cosine ranges between $-1$ and $1$. - Codomain $B = [-2,2]$. - Since $g$ only outputs values in $[-1,1]$, every element of its range is a subset of $B$. - However, values in $B$ outside $[-1,1]$ (like $1.5$) are not achieved by $g$. - Thus, $g$ is **not surjective** onto $B$. 8. **Conclusion:** $g$ is **neither injective nor surjective**, so **not bijective**. ----- ### c) $h(x) = x^2 h$ 9. **Note:** The function $h$ is defined in terms of itself, which is ambiguous. - Assuming a typo and $h(x) = x^2$ instead (which is typical). 10. **Injectivity:** - For $x = 1$ and $x = -1$, $h(1) = 1$, $h(-1) = 1$. - Distinct inputs map to the same output. - So, $h$ is **not injective**. 11. **Surjectivity:** - Codomain is $B = [-2, 2]$. - The range of $h(x) = x^2$ for $x \in [-2,2]$ is $[0,4]$. - Since $4$ is outside $B$, codomain is $[-2,2]$, range is $[0,4]$, outputs like $3$ not in $B$. - But no negative numbers are in the range, so not all elements of $B$ are covered. - Actually, range within $B$ is $[0,2^2] = [0,4]$, but codomain is $[-2,2]$, so outputs greater than 2 are not in codomain. - Typically, the codomain equals $B = [-2,2]$; the function's outputs exceed codomain. - Therefore, $h$ is **not surjective** onto $B$. 12. **Conclusion:** $h$ is **neither injective nor surjective**, so **not bijective**. ----- **Final answers:** - a) $f$ is neither injective nor surjective. - b) $g$ is neither injective nor surjective. - c) $h$ is neither injective nor surjective. Each function is **not bijective**.