Finding Intervals of Increase and Decrease for y = -(4x + 31)²

To determine the intervals of increase and decrease for the function $y = -\left(4x+31\right)^2$, we follow these steps:

• Step 1: Differentiate the function to find its first derivative.
• Step 2: Solve for critical points by setting the derivative equal to zero.
• Step 3: Analyze the sign of the derivative in intervals determined by the critical point.

Now, let's work through each step:

Step 1: Differentiate the function.
The function is $y = -\left(4x+31\right)^2$. Applying the chain rule gives us: $y' = \frac{d}{dx}\left[-(4x+31)^2\right] = -2(4x+31) \cdot \frac{d}{dx}(4x+31) = -2(4x+31) \cdot 4 = -8(4x+31)$ Thus, the derivative is $y' = -8(4x+31)$.

Step 2: Solve for critical points.
Set the derivative equal to zero: $-8(4x+31) = 0$ $4x+31 = 0$ $4x = -31$ $x = -\frac{31}{4} = -7\frac{3}{4}$ This critical point divides the x-axis into two intervals: $x < -7\frac{3}{4}$ and $x > -7\frac{3}{4}$.

Step 3: Analyze the sign of the derivative.
For $x < -7\frac{3}{4}$, say $x = -8$: $y' = -8(4(-8) + 31) = -8(-32 + 31) = -8(-1) = 8$ The derivative is positive, indicating the function is increasing.

For $x > -7\frac{3}{4}$, say $x = 0$: $y' = -8(0 + 31) = -8 \times 31 = -248$ The derivative is negative, indicating the function is decreasing.

Thus, the function is increasing for $x < -7\frac{3}{4}$ and decreasing for $x > -7\frac{3}{4}$.

Therefore, the solution to the problem is $\searrow:x > -7\frac{3}{4} \\\nearrow:x < -7\frac{3}{4}$.