Product Representation: Finding Increasing or Decreasing Domains

To find the intervals where the function $y = (x+1)(7-x)$ is decreasing, we begin by expanding the function:
$y = (x+1)(7-x) = 7x - x^2 + 7 - x = -x^2 + 6x + 7$.

The function is now in the form $y = -x^2 + 6x + 7$.
This is a quadratic function, opening downward because the coefficient of $x^2$ is negative.

Let's find the critical points by taking the derivative and setting it to zero.
The derivative of $y$ is$\ y' = \frac{d}{dx}(-x^2 + 6x + 7) = -2x + 6$.
Solving $-2x + 6 = 0$ gives $x = 3$.

The vertex, $x = 3$, is where the function changes from increasing to decreasing.

To determine the interval where the function is decreasing, consider the derivative:$<br> -2x + 6 < 0$.
Solving gives $-2x < -6$, resulting in $x > 3$.

Therefore, the function is decreasing for $x > 3$.

The correct answer is: $x > 3$

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: $x < 3$

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 $x < -3$.

To find the intervals where the function $y = (x+1)(x+5)$ is increasing, we need to analyze its derivative.

We start by expanding the function: $y = (x+1)(x+5) = x^2 + 6x + 5$.

Next, we find the derivative: $y' = \frac{d}{dx}(x^2 + 6x + 5) = 2x + 6$.

To find where the function is increasing, solve the inequality $2x + 6 > 0$:

• Subtract 6 from both sides: $2x > -6$
• Divide by 2: $x > -3$

This tells us that the function is increasing on the interval $x > -3$.

By analyzing the derivative, the function transitions at $x = -3$, from decreasing (when $x < -3$) to increasing (when $x > -3$).

Therefore, the solution is $x > -3$.

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.

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:

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

To solve this problem, we'll follow these steps:

• Step 1: Identify a simpler form of the equation and its derivative.
• Step 2: Find critical points where the derivative is zero.
• Step 3: Determine signs of the derivative on intervals around critical points.

Let's proceed with these steps:

Step 1: Our function is already in vertex form: $y = \left(2x - 2\frac{1}{4}\right)^2$. It's important to note that this term can be simplified as $y = (2x - 2.25)^2$, identifying the vertex at $x = 1.125$, which is $x = 1\frac{1}{8}$.

Step 2: Differentiate the function with respect to $x$. For $y = (2x - 2.25)^2$, use the chain rule:
$y' = 2(2x - 2.25) \cdot 2 = 4(2x - 2.25) = 8x - 9$.

Step 3: Set the derivative to zero to find critical points:
$8x - 9 = 0$ leads to $x = \frac{9}{8} = 1.125$ or $x = 1\frac{1}{8}$.

The function decreases to the left of this point and increases to the right. Specifically:

• If $x < 1\frac{1}{8}$, $8x - 9 < 0$, so $y$ is decreasing.
• If $x > 1\frac{1}{8}$, $8x - 9 > 0$, so $y$ is increasing.

Therefore, the intervals of increase and decrease are:

$\nearrow:x<1\frac{1}{8}$ (Increasing: to the left of the vertex),

$\searrow:x>1\frac{1}{8}$ (Decreasing: to the right of the vertex).

To solve this problem, we need to determine where the function $y = -(x-16)^2$ is increasing and decreasing.

The function given, $y = -(x-16)^2$, represents a quadratic function with a vertex form of $y = a(x-h)^2 + k$, where $a = -1$, $h = 16$, and $k = 0$. This form shows that the parabola opens downwards because the coefficient $a = -1$ is negative.

The vertex of the parabola, found at $x = 16$, is the point where the function changes its direction. For $x < 16$, since the parabola opens downwards, the function is increasing as it moves toward the vertex. For $x > 16$, the function is decreasing as it moves away from the vertex.

Thus, the intervals are:
- Increasing: $x < 16$
- Decreasing: $x > 16$

The correct solution to the problem, which matches the given answer, is: $\searrow: x > 16 \\ \nearrow: x < 16$.

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 solve this problem, we'll find the intervals of increase and decrease for the function $y = (7x + 3)(5x - 2)$.

First, let's expand the function:

$y = (7x + 3)(5x - 2) = 35x^2 + 15x - 14x - 6 = 35x^2 + x - 6$.

Next, compute the derivative $y'$:

$y' = \frac{d}{dx}(35x^2 + x - 6) = 70x + 1$.

To find critical points, set $y' = 0$:

$70x + 1 = 0$

$70x = -1$

$x = -\frac{1}{70}$.

Now, we need to determine the sign of $y'$ in the intervals around the critical point $x = -\frac{1}{70}$:

• For $x < -\frac{1}{70}$, choose a test point such as $x = -1$:
  $y' = 70(-1) + 1 = -70 + 1 = -69$, so $y' < 0$, indicating that the function is decreasing in this interval.
• For $x > -\frac{1}{70}$, choose a test point such as $x = 0$:
  $y' = 70(0) + 1 = 1$, so $y' > 0$, indicating that the function is increasing in this interval.

Putting it all together, we have:

The function is decreasing on the interval $x < -\frac{1}{70}$ and increasing on the interval $x > -\frac{1}{70}$.

Therefore, the intervals of increase and decrease are:

$\searrow:x < -\frac{1}{70} \\ \nearrow:x > -\frac{1}{70}$.

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:

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

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:
$\searrow: x < 3.35 \\\nearrow: x > 3.35$

To solve the problem of finding intervals of increase and decrease for the given function, follow these steps:

• Step 1: Expand the function.
  We start by expanding $y = \left(2x - \frac{1}{2}\right)\left(x - 2\frac{1}{4}\right)$:
  $y = 2x \cdot x - 2x \cdot 2\frac{1}{4} - \frac{1}{2} \cdot x + \frac{1}{2} \cdot 2\frac{1}{4}$.
  Simplifying, we get:
  $y = 2x^2 - \frac{9}{2}x + \frac{1}{2} \cdot \frac{9}{4}$.
  Thus, $y = 2x^2 - \frac{9}{2}x + \frac{9}{8}$.

• Step 2: Differentiate the function.
  Differentiate $y = 2x^2 - \frac{9}{2}x + \frac{9}{8}$ with respect to $x$:
  $\frac{dy}{dx} = 4x - \frac{9}{2}$.

• Step 3: Find the critical points.
  Set the first derivative equal to zero:
  $4x - \frac{9}{2} = 0$.
  Solving for $x$, we get $4x = \frac{9}{2}$, hence $x = \frac{9}{8}$.

• Step 4: Use the first derivative test.
  Evaluate the sign of $\frac{dy}{dx}$ around the critical point $x = \frac{9}{8}$:
  - For $x < \frac{9}{8}$, choose $x = 1$: $\frac{dy}{dx} = 4(1) - \frac{9}{2} = \frac{8}{2} - \frac{9}{2} = -\frac{1}{2}$ (negative).
  - For $x > \frac{9}{8}$, choose $x = 2$: $\frac{dy}{dx} = 4(2) - \frac{9}{2} = \frac{8}{1} - \frac{9}{2} = \frac{7}{2}$ (positive).
  Thus, the function decreases when $x < \frac{9}{8}$ and increases when $x > \frac{9}{8}$.

Conclusively, the intervals of increase and decrease are:

$\searrow: x < 1\frac{1}{4}, \nearrow: x > 1\frac{1}{4}$.

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 solve this problem, let's follow through the steps:

• Step 1: Differentiate the function $y=(2x-16)^2$ with respect to $x$.

The derivative of $y$ is found using the chain rule. The outer function is $u^2$ with $u = (2x-16)$. Thus, the derivative is:

$y' = 2(2x-16) \cdot \frac{d}{dx}(2x-16)$

This simplifies to:

$y' = 2(2x-16) \cdot 2 = 4(2x-16) = 8x - 64$

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

Set $8x - 64 = 0$:

$8x = 64$

$x = 8$, so $x = 8$ is a critical point.

• Step 3: Analyze each interval determined by this critical point to determine where $y$ is increasing or decreasing.

The number line is split into two intervals by the critical point $x = 8$: $(-∞, 8)$ and $(8, ∞)$.

For $x < 8$ (e.g., $x = 7$):

Evaluate $y'(7) = 8(7) - 64 = 56 - 64 = -8$, so $y'(x) < 0$. Thus, $y$ is decreasing for $x < 8$.

For $x > 8$ (e.g., $x = 9$):

Evaluate $y'(9) = 8(9) - 64 = 72 - 64 = 8$, so $y'(x) > 0$. Thus, $y$ is increasing for $x > 8$.

Therefore, the function $y = (2x-16)^2$ is:
Decreasing on the interval $(-∞, 8)$ and Increasing on the interval $(8, ∞)$.

This corresponds to the correct answer choice:

$\searrow:x<8\\\nearrow:x>8$

To solve this problem, follow these steps:

• Step 1: Expand the function
  We have $y = (3x + 1)(4x - 2)$. Expanding this, we get:

$y = 3x \cdot 4x + 3x \cdot (-2) + 1 \cdot 4x + 1 \cdot (-2)$

$y = 12x^2 - 6x + 4x - 2$

$y = 12x^2 - 2x - 2$

• Step 2: Find the derivative
  The derivative $y'$ of the function $y = 12x^2 - 2x - 2$ is:

$y' = \frac{d}{dx}(12x^2 - 2x - 2)$

$y' = 24x - 2$

• Step 3: Determine critical points
  Set the derivative equal to zero to find critical points:

$24x - 2 = 0$

$24x = 2$

$x = \frac{1}{12}$

• Step 4: Determine intervals of increase and decrease
  Evaluate the sign of $y' = 24x - 2$ on intervals relative to the critical point $x = \frac{1}{12}$:

Test values:

$y'(0) = 24(0) - 2 = -2$ (Negative, so decreasing)

- For $x > \frac{1}{12}$, choose $x = 1$:

$y'(1) = 24(1) - 2 = 22$ (Positive, so increasing)

Therefore, the function is decreasing for $x < \frac{1}{12}$ and increasing for $x > \frac{1}{12}$.

The correct answer is:

$\searrow:x<\frac{1}{12}\\\nearrow:x>\frac{1}{12}$

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 solve this problem, we'll begin by finding the derivative of the given function $y = (4x + 22)^2$ with respect to $x$.

The function can be expanded as:

$y = (4x + 22)^2 = 16x^2 + 2 \cdot 4 \cdot 22 \cdot x + 22^2 = 16x^2 + 176x + 484$

Next, find $\frac{dy}{dx}$ by differentiating:

$\frac{dy}{dx} = \frac{d}{dx}(16x^2 + 176x + 484) = 32x + 176$

Set the derivative equal to zero to find the critical point:

$32x + 176 = 0$

$32x = -176$

$x = -\frac{176}{32}$

$x = -\frac{11}{2}$

$x = -5.5$

This critical point, $x = -5.5$, will be the vertex of the parabola, determining where the function changes from decreasing to increasing.

Now, test the sign of $\frac{dy}{dx} = 32x + 176$ in intervals around the critical point:

• For $x < -5.5$, choose $x = -6$:

$32(-6) + 176 = -192 + 176 = -16$

Since the derivative is negative, the function is decreasing.

• For $x > -5.5$, choose $x = -5$:

$32(-5) + 176 = -160 + 176 = 16$

Since the derivative is positive, the function is increasing.

Therefore, the intervals of increase and decrease are:

$\searrow:x < -5\frac{1}{2}$

$\nearrow:x > -5\frac{1}{2}$

Thus, the correct choice from the given options is:

$\searrow:x < -5\frac{1}{2}$

$\nearrow:x > -5\frac{1}{2}$

To solve this problem, we'll follow these steps:

• Step 1: Expand the original function
• Step 2: Differentiate the function
• Step 3: Find the critical points by setting the derivative to zero
• Step 4: Determine intervals of increase and decrease based on the derivative's sign

Let's explore each step in detail:
Step 1: Expand the function:

The function $y = (x+1)(x+\frac{1}{6})$ expands to:

$y = x^2 + \frac{1}{6}x + x + \frac{1}{6} = x^2 + \frac{7}{6}x + \frac{1}{6}$.

Step 2: Differentiate the function:
The derivative of the quadratic function $y = x^2 + \frac{7}{6}x + \frac{1}{6}$ is:

$y' = 2x + \frac{7}{6}$.

Step 3: Set the derivative equal to zero to find critical points:
Solve $2x + \frac{7}{6} = 0$.

Subtract $\frac{7}{6}$ from both sides: $2x = -\frac{7}{6}$.

Divide both sides by 2 to solve for $x$: