If ⨍ (x)=x-5 ⨍or x≤14x2-9 ⨍or 1<x<2 3x+4 ⨍or x≥2 then ⨍ (2+)-⨍ (2-)=

⨍ (x)=x-5 ⨍or x≤14x2-9 ⨍or 1<x<2 3x+4 ⨍or x≥2 f2+=limx→2+fx=limx→2+3x+4=32+4=10 f2-=limx→2-fx=limx→2+4x2-9=44-9=7 ⨍ (2+)-⨍ (2-)=10-7=3

If ⨍ (x)=x-5 ⨍or x≤14x2-9 ⨍or 1<x<2 3x+4 ⨍or x≥2 then ⨍ (2+)-⨍ (2-)=

⨍ (x)=x-5 ⨍or x≤14x2-9 ⨍or 1<x<2 3x+4 ⨍or x≥2 f2+=limx→2+fx=limx→2+3x+4=32+4=10 f2-=limx→2-fx=limx→2+4x2-9=44-9=7 ⨍ (2+)-⨍ (2-)=10-7=3

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If $⨍ (x)= \{\begin{matrix} x - 5 ⨍or x≤1 \\ 4 x^{2} - 9 ⨍or 1<x<2 \\ 3 x + 4 ⨍or x≥2 \end{matrix}$ then ${⨍ (2}^{+} {)-⨍ (2}^{-})=$

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Detailed Solution

$⨍ (x)= \{\begin{matrix} x - 5 ⨍or x≤1 \\ 4 x^{2} - 9 ⨍or 1<x<2 \\ 3 x + 4 ⨍or x≥2 \end{matrix} f (2^{+}) = \lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} 3x+4=3 (2) +4=10 f (2^{-}) = \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{+}} 4 x^{2} - 9 =4 (4) -9=7 {⨍ (2}^{+} {)-⨍ (2}^{-})=10-7=3$