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For what value of the constant $ c $ is the function $ f $ continuous on $ (-\infty, \infty) $?

$ f(x) = \left\{

\begin{array}{ll}

cx^2 + 2x & \mbox{if $ x < 2 $}\\

x^3 - cx & \mbox{if $ x \ge 2 $}

\end{array} \right.$

$f$ is continuous on $(-\infty, 2)$ and $(2, \infty) .$ Now $\lim _{x \rightarrow 2-} f(x)=\lim _{x \rightarrow 2-}\left(c x^{2}+2 x\right)=4 c+4$ and

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}\left(x^{3}-c x\right)=8-2 c .$ So $f$ is continuous $\Leftrightarrow 4 c+4=8-2 c \Leftrightarrow 6 c=4 \Leftrightarrow c=\frac{2}{3} .$ Thus, for $f$

to be continuous on $(-\infty, \infty), c=\frac{2}{3}$

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consider the piecewise function F of x equals c X squared plus two X. If x is less than two and f of x equals x cubed minus C X if x is greater than or equal to two. Now, in this problem we are to find the value of C in which these functions continue us on negative infinity to infinity. Notes that this function is already concerned was over the interval negative infinity to two. And over the interval truth infinity. This is because the function is defined in ancient herbal as a polynomial. And so we need only show that the function is continuous at X equals two. That is we need to show that the limit as X approaches to of dysfunction exists. Now this exists if The one sided limit limit as X approaches, let's say two from the left of this function is equal to the limit As extra purchased two from the right of dysfunction. Now, the limit as X approaches to you from the left of this function, this is just limit As X approaches to from the left of, we will use the um function C X squared plus two X. Since X approaches to Two from the left, that means X must be less than two. So we do see X squared plus two X. And then from here we want to evaluate the function at two and so we get C times two squared plus two times to that four. C Plus four. And then the limit of the function as X approaches to from the right, this is equal to the limit as X approaches to from the right of the function X cubed minus c. X. We're using this because the Values of x here are those greater than two. And so we have evaluating it to we get to to the third power- C, Times two. That's just 8 -2 C. And because you want this function to be continuous then they're forced to yeah, make the one sided limits equal means we right or C Plus for this should be equal to 8 -2 c. and solving for C, we have four C plus to see this is equal to eight minus four. And so six. See this is equal to four or that C is equal to 4/6 or 2/3. So this is the value of C in which the function is continue was over the interval negative infinity to infinity.