Answer to Question #129376 in Discrete Mathematics for jaya

Question #129376
(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).
1
Expert's answer
2020-08-16T20:31:24-0400

f(x)=x2+1f(x) = x^2 +1

If f:ABf:A→B is a function where HBH⊆B then the inverse image of H under the function f is defined to be the set f1(H)={xA:f(x)H}f^{-1}(H)=\{x∈A:f(x)∈H\}

So we need to find all x for which 5f(x)105 \leq f(x) \leq 10

5x2+1105≤x^2 +1≤10

4x294≤x^2≤9

(x2)(x+2)0(x3)(x+3)0(x-2)(x+2) \geq 0 \cap (x-3)(x+3) \leq 0

The last inequalities give us 3x22x3.−3≤x≤−2 \cup 2≤x≤3.

The answer: f1(H)={xR:(3x2)(2x3)}.f^{−1}(H)=\{x∈R:(−3≤x≤−2) \cup (2≤x≤3)\}.


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