Product Representation: Finding Increasing or Decreasing Domains

First, we need to express the given function in a form that's easy to differentiate:

The original function is $y = (x-6)(x+6)$. Expanding this, we have:

$y = x^2 - 36$.

Next, we'll find the derivative of this quadratic function to determine the intervals where the function is increasing. The derivative will provide the slope of the tangent at any point on the function:

The derivative of $y = x^2 - 36$ is:

$y' = 2x$.

Now, we determine where the derivative $2x$ is positive. A function is increasing where its derivative is positive:

Solve $2x > 0$:

$x > 0$.

This shows that the function is increasing on the interval where $x > 0$.

Therefore, the solution to the problem is that the function is increasing for $x > 0$.

Let's solve the problem step by step:

• **Step 1:** Identify the points $p$ and $q$ from the function $y = (x-4.4)(x-2.3)$. Here, $p = 4.4$ and $q = 2.3$.
• **Step 2:** Calculate the vertex $x$-coordinate using the formula for the midpoint: $x = \frac{p+q}{2} = \frac{4.4 + 2.3}{2}$.
• **Step 3:** Compute the value: $x = \frac{6.7}{2} = 3.35$.
• **Step 4:** Since the quadratic has a positive leading coefficient after expansion (implying it opens upwards), the function decreases on $(-\infty, 3.35)$ and increases on $(3.35, \infty)$.

Therefore, the function is decreasing for $x < 3.35$ and increasing for $x > 3.35$.

The correct choice that matches this conclusion is:
$\searrow: x < 3.35 \\\nearrow: x > 3.35$

Let's solve this problem step-by-step:

The function is given by:

$y = (3x + 1)(1 - 3x)$

We can first find the derivative $y'$ to determine where the function is increasing:

• First, expand the product: $y = 3x + 1 - 9x^2 - 3x$
• Simplify: $y = -9x^2 + 0x + 1$

Now the function looks like this quadratic form $y = -9x^2 + 1$.

Next, compute the derivative:

$\frac{dy}{dx} = -18x$

To find the critical points, set $\frac{dy}{dx} = 0$:

$-18x = 0$

Solving, we find $x = 0$.

Now analyze the sign of $\frac{dy}{dx} = -18x$ around this critical point:

• For $x < 0$, $-18x > 0$. Therefore, the function is increasing.
• For $x > 0$, $-18x < 0$. Therefore, the function is decreasing.

Therefore, the solution is that the function is increasing on the interval:

$x < 0$.

The function given is $y = \left(x - \frac{3}{4}\right)^2$. This is a quadratic function with its vertex (or minimum point) at $x = \frac{3}{4}$.

To find where the function is increasing or decreasing, follow these steps:

• Step 1: Differentiate $y$ with respect to $x$.

$y' = \frac{d}{dx}\left[\left(x - \frac{3}{4}\right)^2\right] = 2\left(x - \frac{3}{4}\right)$

• Step 2: Find the critical points by setting the derivative equal to zero.

$2\left(x - \frac{3}{4}\right) = 0 \quad \Rightarrow \quad x = \frac{3}{4}$

• Step 3: Determine the sign of $y'$ in the intervals divided by the critical point.

For $x < \frac{3}{4}$, $\left(x - \frac{3}{4}\right) < 0$ and hence $y' < 0$, indicating decreasing behavior.

For $x > \frac{3}{4}$, $\left(x - \frac{3}{4}\right) > 0$ and hence $y' > 0$, indicating increasing behavior.

Thus, the function decreases on the interval $(-\infty, \frac{3}{4})$, and increases on the interval $(\frac{3}{4}, \infty)$.

Consequently, the intervals of increase and decrease are:

$\searrow: x > \frac{3}{4} \\ \nearrow: x < \frac{3}{4}$

The function given is $y = (x-4)(-x+6)$. To analyze its behavior, we first convert this into a standard quadratic form by expanding:

$y = (x-4)(-x+6) = -x^2 + 6x + 4x - 24 = -x^2 + 10x - 24$.

The derivative with respect to $x$ of the function $y = -x^2 + 10x - 24$ is $\frac{dy}{dx} = -2x + 10$.

To find critical points, we set the derivative equal to zero:

$-2x + 10 = 0$

Solving for $x$, we find:

$-2x = -10$

$x = 5$.

Next, we test intervals around the critical point $x = 5$:

• For $x < 5$, choose a test point like $x = 0$: $-2(0) + 10 = 10 > 0$, so the derivative is positive, indicating the function is increasing.
• For $x > 5$, choose a test point like $x = 6$: $-2(6) + 10 = -2 < 0$, so the derivative is negative, indicating the function is decreasing.

Therefore, the function $y = (x-4)(-x+6)$ is decreasing for $x > 5$.

Thus, the correct answer is $x > 5$.

The function $y = (x+1)(x+5)$ is in intercept form, and we can start by expanding it:

$y = x^2 + 6x + 5$.

We take the derivative of the quadratic function with respect to $x$ to find the critical points:

$y' = \frac{d}{dx}(x^2 + 6x + 5) = 2x + 6$.

Set the derivative equal to zero to find any critical points:

$2x + 6 = 0$.
Solving for $x$, we get $2x = -6$ or $x = -3$.

This critical point, $x = -3$, will help us break the number line into intervals to test whether the derivative is positive or negative.

We examine intervals to determine where the function is decreasing by using test points:

• Interval $x < -3$: Choose a test point (e.g., $x = -4$)
  $y'(-4) = 2(-4) + 6 = -8 + 6 = -2$ (negative)
• Interval $x > -3$: Choose a test point (e.g., $x = 0$)
  $y'(0) = 2(0) + 6 = 6$ (positive)

For $x < -3$, $y'$ is negative, indicating the function is decreasing in this interval.

Therefore, the interval where the function is decreasing is $x < -3$.

The intervals of increase and decrease of the function are:
Decreasing on $x < -\frac{1}{70}$
Increasing on $x > -\frac{1}{70}$.

Therefore, choice 3 is correct.

To determine the decreasing intervals of the function $y = (4x + 8)(-x + 2)$, we follow these steps:

• Step 1: Expand the function.
• Step 2: Compute its derivative.
• Step 3: Find where the derivative is negative.

Step 1: Expand the Function
First, let's expand $y = (4x + 8)(-x + 2)$:

$y = 4x(-x) + 4x(2) + 8(-x) + 8(2)$

$y = -4x^2 + 8x - 8x + 16$

Simplifying, we have $y = -4x^2 + 16$.

Step 2: Compute the Derivative
The derivative of $y$ with respect to $x$ is:

$y' = \frac{d}{dx}(-4x^2 + 16) = -8x$.

Step 3: Find where the Derivative is Negative
We need to solve for $y' < 0$:

$-8x < 0$

This implies $x > 0$.

Therefore, the function $y = (4x + 8)(-x + 2)$ is decreasing on the interval $x > 0$.

To determine the intervals of increase and decrease, follow these steps:

• Step 1: Simplify the function. The given function is $y = -(4x + 32)^2$, a quadratic in terms of $x$.
• Step 2: Recognize that this is a downward-facing parabola because the coefficient of $(4x + 32)^2$ is negative.
• Step 3: Find the vertex or the critical point of the parabola. The function is in the form $y = -a(x + h)^2 + k$. Here, $a = 4$, $h = -8$, and $k = 0$.
• Step 4: The vertex occurs at $x = -8$. The function is symmetrical about this point.
• Step 5: Since it's a downwards-opening parabola, the function increases on the left of the vertex and decreases on the right of the vertex.
• Step 6: Thus, the function decreases ($\searrow$) for $x < -8$ and increases ($\nearrow$) for $x > -8$.

Consequently, the intervals of increase and decrease are:
Decreasing: $x < -8$
Increasing: $x > -8$

Therefore, $\searrow: x<-8 \\ \nearrow: x>-8$ is the correct answer.

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 $\searrow:x > -7\frac{3}{4} \\\nearrow:x < -7\frac{3}{4}$.

To determine the intervals where the function $y = (x+10)(x-8)$ is increasing, we will follow these steps:

• Step 1: Expand the expression for clarity.
• Step 2: Find the derivative $y'$ of the expanded function.
• Step 3: Determine where $y' > 0$ to identify increasing intervals.

Step 1: Expand the function $y = (x+10)(x-8)$.

Expanding gives: $y = x^2 + 2x - 80$.

Step 2: Find the derivative $y'$ of $y = x^2 + 2x - 80$.

$y' = \frac{d}{dx}(x^2 + 2x - 80) = 2x + 2$.

Step 3: Find where the function is increasing by solving $y' > 0$.

$2x + 2 > 0$.

Solve for $x$:

$2x > -2$

$x > -1$.

Thus, the function is increasing for $x > -1$.

Therefore, the solution to the problem is $x > -1$.

To determine where the function $y = (3x+3)(9-x)$ is increasing, we will use the following steps:

• Step 1: Expand and simplify the quadratic expression.
• Step 2: Find the derivative of the function.
• Step 3: Determine where the derivative is positive.

Let's go through these steps:

Step 1: Expand and simplify the quadratic expression:
The function given is $y = (3x + 3)(9 - x)$.

We expand this expression:
$y = 3x \cdot 9 + 3x \cdot (-x) + 3 \cdot 9 + 3 \cdot (-x)$.
This simplifies to:
$y = 27x - 3x^2 + 27 - 3x$.
Combining like terms, we get the quadratic equation:
$y = -3x^2 + 24x + 27$.

Step 2: Find the derivative of the function:
The quadratic equation found is $y = -3x^2 + 24x + 27$.
Taking the derivative, we have:
$y' = \frac{d}{dx}(-3x^2 + 24x + 27) = -6x + 24$.

Step 3: Determine where the derivative is positive:
To find where the function is increasing, solve the inequality:
$y' = -6x + 24 > 0$.
This simplifies to:
$-6x > -24$.
Dividing both sides by -6 (and remembering to reverse the inequality sign) gives:
$x < 4$.

Thus, the function is increasing on the interval where $x < 4$.

Therefore, the solution to the problem is $x < 4$..

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:

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

To determine where the function $y=\left(\frac{1}{3}x+\frac{1}{2}\right)^2$ is increasing or decreasing, we need to first find its derivative.

Let's compute the derivative $y'$ of the function:

$y = \left(\frac{1}{3}x + \frac{1}{2}\right)^2$.

Using the chain rule, let $u = \frac{1}{3}x + \frac{1}{2}$, then $y = u^2$.

The derivative of $u^2$ with respect to $x$ is $2u \cdot \frac{du}{dx}$.

Now, find $\frac{du}{dx}$:

$\frac{du}{dx} = \frac{1}{3}$.

Thus, the derivative of the function is:

$y' = 2 \left( \frac{1}{3}x + \frac{1}{2} \right) \cdot \frac{1}{3} = \frac{2}{3} \left( \frac{1}{3}x + \frac{1}{2} \right)$.

Set this derivative to zero to find critical points:

$\frac{2}{3} \left( \frac{1}{3}x + \frac{1}{2} \right) = 0$.

Simplify to find $x$:

$\frac{1}{3}x + \frac{1}{2} = 0$.

$\frac{1}{3}x = -\frac{1}{2}$.

$x = -\frac{1}{2} \times 3$.

$x = -\frac{3}{2}$.

The critical point is at $x = -\frac{3}{2}$.

To determine the nature of intervals around this critical point, test $y'$ on intervals around $x = -\frac{3}{2}$.

- For $x < -\frac{3}{2}$: Choose $x = -2$.
$y' = \frac{2}{3} \left( \frac{1}{3}(-2) + \frac{1}{2} \right) = \frac{2}{3}(-\frac{2}{3} + \frac{1}{2}) < 0$.
$y'$ is negative, so $y$ is decreasing.

- For $x > -\frac{3}{2}$: Choose $x = 0$.
$y' = \frac{2}{3} \left( \frac{1}{3}(0) + \frac{1}{2} \right) = \frac{2}{3} \times \frac{1}{2} > 0$.
$y'$ is positive, so $y$ is increasing.

Thus, the function decreases on $x < -\frac{3}{2}$ and increases on $x > -\frac{3}{2}$.

The intervals of increase and decrease are $\searrow: x > -1\frac{1}{2}$ and $\nearrow: x < -1\frac{1}{2}$.

Analyzing the multiple-choice answers, the correct one matches choice 3.

To determine where the function $y = (x+1)(7-x)$ is increasing, follow these steps:

• Step 1: Expand the function. $y = (x+1)(7-x) = 7x - x^2 + 7 - x = -x^2 + 6x + 7$
• Step 2: Differentiate the function. $\frac{dy}{dx} = \frac{d}{dx}(-x^2 + 6x + 7) = -2x + 6$
• Step 3: Identify the intervals where the derivative is positive. $-2x + 6 > 0$
• Step 4: Solve the inequality. $-2x + 6 > 0$ $6 > 2x$ $3 > x$ Thus, the function is increasing for $x < 3$.

Therefore, the function $y = (x+1)(7-x)$ is increasing on the interval $x < 3$.

Matching this result with the given choices, the correct choice is:

Choice 3: $x < 3$

To determine where the function $y = (x-4)(x+2)$ is decreasing, we will first convert the product form to standard form:

Step 1: Expand the function: $y = x^2 - 4x + 2x - 8 = x^2 - 2x - 8$

Step 2: Differentiate the function with respect to $x$ to find the derivative $y'$: $y' = \frac{d}{dx}(x^2 - 2x - 8) = 2x - 2$

Step 3: Determine where the derivative is negative: $2x - 2 < 0$

Step 4: Solve for $x$: $2x < 2$ $x < 1$

Therefore, the function $y = (x-4)(x+2)$ is decreasing for $x < 1$.

This corresponds to choice 2: $x<1$

To determine where the function $y = (x-6)(x+6)$ is decreasing, we analyze the quadratic function in its factored form.

Step 1: Identify the roots and the vertex.

• The roots of the function are $x = 6$ and $x = -6$.
• The vertex is exactly halfway between these roots, located at $x = \frac{6 + (-6)}{2} = 0$.

Step 2: Determine the behavior on each side of the vertex.

• This function represents a standard upward-opening parabola because it can be rewritten as $y = x^2 - 36$, which has a positive leading coefficient.
• The parabola decreases as it approaches the vertex from the left and increases as it moves away to the right. Thus, the function is decreasing for values of $x$ less than the vertex, $x = 0$.

Step 3: State the interval where the function is decreasing.

The function is decreasing on the interval $x < 0$.

The correct solution to the problem, where the function is decreasing, is $x < 0$.

To determine where the function $y = (x-9)(5-x)$ is increasing, we follow these steps:

• **Step 1: Expand the expression**: Start by expanding the given function:
  $y = (x-9)(5-x) = x \cdot 5 - x \cdot x - 9 \cdot 5 + 9 \cdot x = 5x - x^2 - 45 + 9x$.
  Simplifying gives: $y = -x^2 + 14x - 45$.
• **Step 2: Differentiate to find $y'$**: Compute the first derivative with respect to $x$:
  $y' = \frac{d}{dx} (-x^2 + 14x - 45) = -2x + 14$.
• **Step 3: Solve $y' = 0$ for critical points**:
  Set $y'$ equal to zero:
  $-2x + 14 = 0$,
  $2x = 14$,
  $x = 7$.
• **Step 4: Test intervals using the derivative**:
  We analyze the sign of the derivative $y' = -2x + 14$ on intervals determined by the critical point $x = 7$.
• For $x < 7$, choose $x = 6$, then $y' = -2(6) + 14 = -12 + 14 = 2$, which is positive. Hence, the function is increasing here.
• For $x > 7$, choose $x = 8$, then $y' = -2(8) + 14 = -16 + 14 = -2$, which is negative. Hence, the function is decreasing here.

**Conclusion**: The function $y = (x-9)(5-x)$ is increasing on the interval $-\infty < x < 7$.

This matches the correct answer choice (2):

$x<7$

To find the intervals of increase and decrease for the function $y = (5x - 1)^2$, follow these steps:

• Step 1: Differentiate the function $y = (5x - 1)^2$.
• Step 2: Set the derivative equal to zero to find critical points.
• Step 3: Test intervals around the critical point to determine where the function is increasing or decreasing.

Now, let's work through each step.

Step 1: Differentiate the function.
The given function is $y = (5x - 1)^2$. Using the chain rule, the derivative $y'$ is:

$y' = 2(5x - 1) \cdot 5 = 10(5x - 1) = 50x - 10$

Step 2: Find critical points.
Set $y' = 0$:

$50x - 10 = 0$

Solving for $x$, we get:

$50x = 10 \quad \Rightarrow \quad x = \frac{1}{5}$

Step 3: Determine intervals of increase and decrease.
Test intervals around the critical point $x = \frac{1}{5}$.

• For $x < \frac{1}{5}$ (e.g., $x = 0$), $y' = 50(0) - 10 = -10$. Therefore, $y' < 0$, and the function is decreasing.
• For $x > \frac{1}{5}$ (e.g., $x = 1$), $y' = 50(1) - 10 = 40$. Therefore, $y' > 0$, and the function is increasing.

Thus, the function is decreasing on the interval $x < \frac{1}{5}$ and increasing on the interval $x > \frac{1}{5}$.

Therefore, the correct choice is:

$\searrow: x < \frac{1}{5} \\ \nearrow: x > \frac{1}{5}$

To find the intervals of increase and decrease for the function $y = (5x - 1)\left(4x - \frac{1}{4}\right)$, we perform the following steps:

• Step 1: Differentiate the function using the product rule.
• Step 2: Find the critical points by setting the derivative to zero.
• Step 3: Determine the intervals where the derivative is positive or negative to infer increasing and decreasing behavior.

Now, let's work through each step in detail:

Step 1: Differentiate the function.
Using the product rule, consider $u = 5x - 1$ and $v = 4x - \frac{1}{4}$. The derivative of the function is:

$y' = \frac{d}{dx}\left[(5x - 1)\left(4x - \frac{1}{4}\right)\right] = (5x - 1)\frac{d}{dx}\left(4x - \frac{1}{4}\right) + \left(4x - \frac{1}{4}\right)\frac{d}{dx}(5x - 1)$

$= (5x - 1) \cdot 4 + \left(4x - \frac{1}{4}\right) \cdot 5 = 20x - 4 + 20x - \frac{5}{4}$

$= 40x - \frac{21}{4}$

Step 2: Find the critical points.
Set the derivative to zero:

$40x - \frac{21}{4} = 0$

Solving for $x$, multiply both sides by 4 to clear the fraction:

$160x - 21 = 0$

$x = \frac{21}{160}$

Step 3: Analyze the sign of the derivative around the critical point to determine increasing or decreasing intervals.
Choose a test point in each interval defined by the critical point $x = \frac{21}{160}$.

• For $x < \frac{21}{160}$, choose $x = 0$ and check the sign of $y' = 40(0) - \frac{21}{4} = -\frac{21}{4}$ which is negative, indicating $y$ is decreasing.
• For $x > \frac{21}{160}$, choose $x = 1$ and check the sign of $y' = 40(1) - \frac{21}{4} = \frac{139}{4}$ which is positive, indicating $y$ is increasing.

Thus, the function decreases for $x < \frac{21}{160}$ and increases for $x > \frac{21}{160}$.

Therefore, the intervals are $\searrow:x<\frac{21}{160}$ and $\nearrow:x>\frac{21}{160}$.

The correct choice is:

$\searrow:x<\frac{21}{160}\\\nearrow:x>\frac{21}{160}$