Finding Intervals of Increase and Decrease for y = (4x+16)²

To determine where the function $y = (4x + 16)^2$ is increasing or decreasing, let's analyze its derivative:

Step 1: Differentiate the function.

The function $y = (4x + 16)^2$ is of the form $(u(x))^2$ where $u(x) = 4x + 16$. The derivative of $y$ with respect to $x$ is:

$y' = 2u(x) \cdot u'(x)$

Here, $u'(x) = 4$, so the derivative is:

$y' = 2(4x + 16) \cdot 4 = 8(4x + 16) = 32x + 128$.

Step 2: Find the critical points.

Set the derivative equal to zero and solve for $x$:

$32x + 128 = 0$

$32x = -128$

$x = -4$.

Step 3: Determine the sign of the derivative around the critical point $x = -4$.

• Choose a test point less than $-4$, for example $x = -5$. Then, $y' = 32(-5) + 128 = -32$, which is negative, indicating the function is decreasing.
• Choose a test point greater than $-4$, for example $x = 0$. Then, $y' = 32(0) + 128 = 128$, which is positive, indicating the function is increasing.

Therefore, the function is decreasing on the interval $x < -4$ and increasing on the interval $x > -4$.

The correct answer choice matches these findings:

$\searrow:x<-4\\\nearrow:x>-4$