Let A = { x ∈ R : x 2 − 5 x + 4 ≤ 0 } , B = (3 , 5), and C = (3 , 4]. Show that A ∩ B = C . (Recall, you must show two separate statements: A ∩ B ⊆ C and C ⊆ A ∩ B . Hint, you may benefit from factoring the polynomials that appear in the sets.)
Accepted Solution
A:
Answer:You can proceed as follows:Step-by-step explanation:First solve the quadratic inequality [tex]x^{2}-5x+4\leq 0[/tex]. To do that, factorize, then we have that [tex](x-4)(x-1)\leq 0[/tex]. This implies that [tex]x-1\leq 0\, \text{and}\, x-4\geq 0[/tex]or [tex]x-1 \geq 0\, \text{and}\, x-4\leq 0[/tex]In the first case the solution is the empty set [tex]\emptyset[/tex]. In the second case the solution is the interval [tex]1\leq x \leq 4[/tex]. Now we have that[tex]A=[1,4][/tex][tex]B=(3,5)[/tex][tex]C=(3,4][/tex].To show that [tex]A\cap B\subseteq C[/tex] consider [tex]x\in A\cap B[/tex]. Then [tex]1\leq x \leq 4\, \text{and}\, 1<x<5[/tex], this implies that [tex]3<x\leq 4[/tex], then [tex]x\in C[/tex]. Now, to show that [tex]C\subseteq A\cap B[/tex] consider [tex]x\in C[/tex], then [tex]3<x\leq 4[/tex], then [tex]1\leq x \leq 4\, \text{and}\, 3<x<5[/tex], then [tex]x\in [1,4] \, \text{and}\, x\in (3,5)[/tex], this implies that [tex]x\in A\cap B[/tex].Observe the image below.