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converting1

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[tex] { x | x^2 - 5x + 6 < 0 } [/tex]

I got the inf and sup to be 2 and 3 respectively

[tex] { x^2 - 5x + 6 | x \in ℝ} [/tex]

here I was rather confused what this is saying. I'm assuming it's taking about the graph x^2 - 5x + 6 and assumed this was between [-1/4, ∞] so inf = -1/4 and sup does not exist as it is not bounded from above.

[tex] {x | x^2 + 1 = 0 } [/tex]

as I'm in a real analysis class, there isn't a real number such that x^2 + 1 = 0, so inf and sup do not exist

could anyone check my answers and if my reasoning is correct, especially for the second one please

I don't understand why the curly brackets are not showing, but there should be curly brackets around all above in tex