f(x)=x2+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≤f(x)≤10
5≤x2+1≤10
4≤x2≤9
(x−2)(x+2)≥0∩(x−3)(x+3)≤0
The last inequalities give us −3≤x≤−2∪2≤x≤3.
The answer: f−1(H)={x∈R:(−3≤x≤−2)∪(2≤x≤3)}.
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