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Find a constant 'k' such that..
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...the function is continuous. \[f(x)\] \[=3x+2\] for \[x <2\] \[=x^{2}+k \] for \[x \ge2\]
Well, let's find the limit the 'lower' half of the function (x < 2) is approaching when x approaches 2: $$ \lim_{x \to 2^-} 3x + 2 = 3 \cdot 2 + 2 = 6 + 2 = 8 $$ Means, the function gets closer and closer to 8 as x grows closer and closer to 2. So we need our point at x=2 to be 8. $$ f(2) = 8 \\ 2^2 + k = 8 \\ 4 + k = 8 \\ k = 4 $$
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