Question 6 (5 points) \( \checkmark \) Saved For what value of the constant c is the function \( f(\mathrm{x}) \) continuous on the interval \( (-\infty, \infty) \) where \[ f(x)=\left\{\begin{array}{ll}c x^{2}+3 & \text { if } x \leq-1 \\ 4 c x-6 & \text { if } x>-1\end{array}\right. \] If the answer in infinity, you can type in INF in the blank. If the answer is negative infinity, you can type -INF. If the answer is does not exist, you can type in DNE. Otherwise, write the number as an integer or reduced fraction.

To ensure the function \( f(x) \) is continuous on \( (-\infty, \infty) \), we need to make sure there is no discontinuity at \( x = -1 \), where the definition of the function changes. Given: \[ f(x) = \begin{cases} c x^{2} + 3 & \text{if } x \leq -1 \\ 4c x - 6 & \text{if } x > -1 \end{cases} \] 1. **Evaluate \( f(x) \) from the left at \( x = -1 \)**: \[ f(-1) = c(-1)^2 + 3 = c + 3 \] 2. **Evaluate \( f(x) \) from the right at \( x = -1 \)**: \[ \lim_{x \to -1^+} f(x) = 4c(-1) - 6 = -4c - 6 \] 3. **Set the left-hand limit equal to the right-hand limit for continuity**: \[ c + 3 = -4c - 6 \] \[ 5c = -9 \implies c = -\frac{9}{5} \] **Final Answer:** \(-\dfrac{9}{5}\)