(a) (i) Let f : R → R be defined by the equation f(x) x 2 + 1. Let H ⊆ R and H = { y ∈ R : 5 ≤ y ≤ 10}. Then determine the inverse image f −1 (H).

$f(x) = x^2 +1$

If $f:A→B$ is a function where $H⊆B$ then the inverse image of H under the function f is defined to be the set $f^{-1}(H)=\{x∈A:f(x)∈H\}$

So we need to find all x for which $5 \leq f(x) \leq 10$

$5≤x^2 +1≤10$

$4≤x^2≤9$

$(x-2)(x+2) \geq 0 \cap (x-3)(x+3) \leq 0$

The last inequalities give us $−3≤x≤−2 \cup 2≤x≤3.$

The answer: $f^{−1}(H)=\{x∈R:(−3≤x≤−2) \cup (2≤x≤3)\}.$