1. **Problem statement:** Given two sets $A = B = \{x \mid -2 \leq x \leq 2\}$, determine whether each function is injective, surjective, or bijective. 2. **Function a) $f(x) = |x|$** - **Injective (one-to-one)** means no two different inputs map to the same output. - For $f(x) = |x|$, observe that $f(-1) = 1$ and $f(1) = 1$, so two different inputs give the same output. - Therefore, $f$ is **not injective**. - **Surjective (onto)** means every element in the codomain $B$ has a preimage in $A$. - Since $f(x)=|x|$ maps $A$ to values in $[0,2]$, it does not cover negative values in $B=[-2,2]$. - So, $f$ is **not surjective** onto $B$. - Since $f$ is neither injective nor surjective, it is **not bijective**. 3. **Function b) $g(x) = \cos\left(\frac{\pi x}{2}\right)$** - Injectivity: check if distinct $x$ values in $[-2,2]$ can produce the same output. - Note $g(-2) = \cos(-\pi) = -1$ and $g(2) = \cos(\pi) = -1$. - Different inputs yield the same output, so $g$ is **not injective**. - Surjectivity: The cosine function over $[-2,2]$ goes from $-1$ to $1$ continuously and includes all values in between. - Hence, $g$ is **surjective** onto $B=[-2,2]$ only if the codomain is restricted to $[-1,1]$. - Since the codomain is $[-2,2]$, and $g(x)$ gives values only in $[-1,1]$, it is **not surjective** onto $B$. - Therefore, $g$ is **not bijective**. 4. **Function c) $h(x) = x^2h$** - Assuming the notation means $h(x) = x^2$ with codomain $B=[-2,2]$. - Injectivity: $h(-1) = 1$ and $h(1)=1$, so $h$ is **not injective**. - Surjectivity: the range of $h$ on $[-2,2]$ is $[0,4]$, but codomain is $[-2,2]$. - Values less than 0 (negative) in codomain are not attained. - Also $4 \notin B$. - So $h$ is **not surjective** onto $B$. - Thus, $h$ is **not bijective**. **Final summary:** - $f$ is neither injective nor surjective. - $g$ is surjective onto $[-1,1]$ but not onto $[-2,2]$, and not injective. - $h$ is neither injective nor surjective.