Q.The function $f(x) = \begin{cases} \dfrac{x^2 + 2x + 1}{x^2 - 8x + 12} & x \ne 3,4 \ 0 & x = 3 \ 1 & x = 4 \end{cases}$ is continuous at

x = 3 only

x = 4 only

both

neither

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.The function f(x)={x2+2x+1x2−8x+12x≠3,4 0x=3 1x=4f(x) = \begin{cases} \dfrac{x^2 + 2x + 1}{x^2 - 8x + 12} & x \ne 3,4 \ 0 & x = 3 \ 1 & x = 4 \end{cases}f(x)={x2−8x+12x2+2x+1​​x=3,4 0​x=3 1​x=4​ is continuous at