\begin{tabular}{l} Find the critical points and determine if the function is increasing or decreasing on the given intervals. \\ \( y=6 x^{4}+8 x^{3} \) \\ Left critical point: \( c_{1}= \) \\ Right critical point: \( c_{2}= \) \\ The function is: \\ \begin{tabular}{l} ? \\ \hline\( ? \) \\ \hline\( ? \)\end{tabular} \\ \hline on \( \left(-\infty, c_{1}\right) \)\end{tabular}

We begin with the function \[ y = 6x^4 + 8x^3. \] 1. Compute the derivative: \[ y' = \frac{d}{dx}\left(6x^4 + 8x^3\right) = 24x^3 + 24x^2. \] 2. Factor the derivative: \[ y' = 24x^2(x+1). \] 3. Set the derivative equal to zero to find the critical points: \[ 24x^2(x+1) = 0. \] This gives: \[ x^2 = 0 \quad \text{or} \quad x+1=0. \] Thus: \[ x = 0 \quad \text{or} \quad x = -1. \] We assign \[ c_{1} = -1 \quad \text{(left critical point)} \quad \text{and} \quad c_{2} = 0 \quad \text{(right critical point)}. \] 4. Determine the sign of the derivative on each interval. - **Interval \(\left(-\infty, c_1\right)=\left(-\infty, -1\right)\):** Choose \(x = -2\): \[ y'(-2) = 24(-2)^2(-2+1) = 24(4)(-1) = -96. \] Since \(y'(-2) < 0\), the function is **decreasing** on \(\left(-\infty, -1\right)\). - **Interval \(\left(c_1, c_2\right)=(-1, 0)\):** Choose \(x = -0.5\): \[ y'(-0.5) = 24(-0.5)^2(-0.5+1) = 24(0.25)(0.5) = 3. \] Since \(y'(-0.5) > 0\), the function is **increasing** on \(\left(-1, 0\right)\). - **Interval \(\left(c_2, \infty\right)=(0, \infty)\):** Choose \(x = 1\): \[ y'(1) = 24(1)^2(1+1) = 24(1)(2) = 48. \] Since \(y'(1) > 0\), the function is **increasing** on \(\left(0, \infty\right)\). 5. Final answers: \[ \begin{array}{l} \text{Left critical point: } c_{1} = -1 \\ \text{Right critical point: } c_{2} = 0 \\ \text{The function is:} \\ \quad \text{Decreasing on } \left(-\infty, -1\right) \\ \quad \text{Increasing on } \left(-1, 0\right) \\ \quad \text{Increasing on } \left(0, \infty\right) \end{array} \]