For what value of the constant $$ c $$ is the function $$ f $$ continuous on $$ (-\infty, \infty) $$ ? $$ f(x)=\left\{\begin{array}{ll}c x^{2}+2 x & \text { if } x

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For what value of the constant $$ c $$ is the function $$ f $$ continuous on $$ (-\infty, \infty) $$ ? $$ f(x)=\left\{\begin{array}{ll}c x^{2}+2 x & \text { if } x<3 \ x^{3}-c x & \text { if } x \geq 3\end{array}\right. $$ Need Help? Watch it SUBMIT ANSWER

UpStudy AI · Accepted Answer

To ensure that the function $$ f $$ is continuous on $$ (-\infty, \infty) $$, we need to make sure that there is no break or jump at the point where the definition of the function changes, which is at $$ x = 3 $$.

The function is defined as:
$$ 
f(x) = \begin{cases} 
c x^{2} + 2 x & \text{if } x < 3 \
x^{3} - c x & \text{if } x \geq 3 
\end{cases} 
$$

1. **Continuity within Each Piece:**
   - For $$ x < 3 $$, $$ f(x) = c x^{2} + 2 x $$ is a polynomial, which is continuous for all real numbers.
   - For $$ x \geq 3 $$, $$ f(x) = x^{3} - c x $$ is also a polynomial and thus continuous for all real numbers.

2. **Continuity at $$ x = 3 $$:**
   - The left-hand limit as $$ x $$ approaches 3 from below is:
     $$
     \lim_{x \to 3^-} f(x) = c(3)^2 + 2(3) = 9c + 6
     $$
   - The value of the function at $$ x = 3 $$ (from the right) is:
     $$
     f(3) = 3^3 - c(3) = 27 - 3c
     $$
   - For continuity at $$ x = 3 $$, these two expressions must be equal:
     $$
     9c + 6 = 27 - 3c
     $$
   - Solving for $$ c $$:
     $$
     9c + 3c = 27 - 6 \
     12c = 21 \
     c = \frac{21}{12} = \frac{7}{4}
     $$

**Final Answer:**  
The constant $$ c $$ must be $$ \frac{7}{4} $$ for $$ f $$ to be continuous on $$ (-\infty, \infty) $$.
$$ c = \frac{7}{4} $$

For what value of the constant \( c \) is the function \( f \) continuous on \( (-\infty, \infty) \) ? \[ f(x)=\left\{\begin{array}{ll}c x^{2}+2 x & \text { if } x<3 \\ x^{3}-c x & \text { if } x \geq 3\end{array}\right. \] Need Help? Watch it SUBMIT ANSWER

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